ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: |ll|lll| on April 25, 2010, 11:09:47 pm

Title: cherylim23's methods question thread :)
Post by: |ll|lll| on April 25, 2010, 11:09:47 pm: Hi, I don't understand why y is halved and x remains the same when y = x^3 is dilated by a factor of half from the x-axis.
Got that from heinemann textbook.
Title: Re: cherylim23's methods question thread :)
Post by: the.watchman on April 25, 2010, 11:13:46 pm: The process of dilating by a factor of 'k' from the y-axis is replacing x with $\frac{x}{k}$

So, in this case, $y=(\frac{x}{1/2})^3= 8x^3$
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on April 25, 2010, 11:16:29 pm: Oops, sorry. I meant x-axis.
The qn stated: What happens to the ht and width of this portion of the slide if the function is dilated by a factor of 1r2
The ht supposedly was halved and the width stays the same.
Title: Re: cherylim23's methods question thread :)
Post by: the.watchman on April 25, 2010, 11:21:15 pm: So, for dilation by 1/2 from x-axis, y should be replaced by y/(1/2)=2y

If it helps, draw the old and new graphs and compare them
The new graph should look squished vertically, but it can be interpreted as a stretch horizontally as well :)
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on April 25, 2010, 11:38:22 pm: Yeah, I did them on my CAS. However, for the original one, it has a y-intercept at (0,12). After dilation, the y-intercept is (0,10).
(Graphs are: -4r125 (x-5)^3 + 8 and -2r125 (x-5)^3 + 8
Title: Re: cherylim23's methods question thread :)
Post by: the.watchman on April 26, 2010, 08:34:50 am: Quote from: cherylim23 on April 25, 2010, 11:38:22 pm
Yeah, I did them on my CAS. However, for the original one, it has a y-intercept at (0,12). After dilation, the y-intercept is (0,10).
(Graphs are: -4r125 (x-5)^3 + 8 and -2r125 (x-5)^3 + 8

In this case, the intercept is affected by the translation (up 8 units) AFTER the dilation, so the intercept does not double
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on April 26, 2010, 09:04:02 am: So the answer is probably wrong? You mean the height (y intercept) decreases but the width stays the same?
I'm not too sure about the width either...
Title: Re: cherylim23's methods question thread :)
Post by: the.watchman on April 26, 2010, 09:12:19 am: I don't think you understood what I meant :)
There is a translation up eight units, so the 'new' line from which the dilations occur is y=8
Considering that, isn't the second intercept double the first, if you are talking about the distance from the line y=8?
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on April 26, 2010, 11:20:27 am: Oh yeah, I get what you mean :)
However, for the question, i think the dilation was after the translation...
Oh well, nevermind. I have another question :D

For y=(x-a)^3(x-b), where a and b > 0,
as x approaches infinity, y approaches infinity. (i.e. x --> i, y --> i)

how about, when x approaches negative infinity, what happens to y? (i.e. x --> -i, y --> ?)

Also, the y-intercept is at a^3b right?
Title: Re: cherylim23's methods question thread :)
Post by: the.watchman on April 26, 2010, 11:26:51 am: Considering the type of graph, it is a positive quartic (power 4, positive co-efficient)
So as $x \to -\infty, y \to \infty$
Title: Re: cherylim23's methods question thread :)
Post by: brightsky on April 26, 2010, 12:37:43 pm: Quote from: cherylim23 on April 26, 2010, 11:20:27 am
For y=(x-a)^3(x-b), where a and b > 0,
Also, the y-intercept is at a^3b right?

Yep.
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on April 26, 2010, 01:01:22 pm: Quote from: the.watchman on April 26, 2010, 11:26:51 am
Considering the type of graph, it is a positive quartic (power 4, positive co-efficient)
So as $x \to -\infty, y \to \infty$

okay, so you just ignore the minimum part where x < 0?
Title: Re: cherylim23's methods question thread :)
Post by: the.watchman on April 26, 2010, 01:03:11 pm: Quote from: cherylim23 on April 26, 2010, 01:01:22 pm
Quote from: the.watchman on April 26, 2010, 11:26:51 am
Considering the type of graph, it is a positive quartic (power 4, positive co-efficient)
So as $x \to -\infty, y \to \infty$

okay, so you just ignore the minimum part where x < 0?

What do you mean? :)

Another way to work it out is to think of the y-value, if you let x be a very big number
Title: Re: cherylim23's methods question thread :)
Post by: brightsky on April 26, 2010, 01:52:50 pm: Quote from: cherylim23 on April 26, 2010, 01:01:22 pm
Quote from: the.watchman on April 26, 2010, 11:26:51 am
Considering the type of graph, it is a positive quartic (power 4, positive co-efficient)
So as $x \to -\infty, y \to \infty$

okay, so you just ignore the minimum part where x < 0?

You can also think of it this way.

$y = (x-a)^3 (x-b)$

If you substitute in $x = -\infty$ , in other words, a very big negative number, then $x - a$ would be a very big negative number and $(x - a)^3$ would also be a very big negative number.

Now as for $x - b$ , when you substitute in x as a very big negative number, $x - b$ would also be a very big negative number.

Now the equation turns out to be $y = \text{Very big negative number} \times \text{Very big negative number}$

$\implies y = \text{Very big POSITIVE number}$

$\implies y = \infty$

Hope this helps. :)
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on April 26, 2010, 02:30:59 pm: Yeah, thanks. :D
I understand both replies.
But if you plot the graph, between a < x < b (assuming b > a), y decreases instead of increasing.
So do I just ignore the part for a < x < b?
Title: Re: cherylim23's methods question thread :)
Post by: the.watchman on April 26, 2010, 02:44:28 pm: That is because, as indicated by the eqn, the graph has a stationary point of inflexion at (a,0), so the graph would keep decreasing until another stationary point, definitely between a and b
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on April 26, 2010, 09:34:21 pm: Another question (which I kinda not get)
Find the value of k for which a - 3b is a factor of a^4 - (7a^2)(b^2) + k(b^4).
Hence, for this value of k, factorise a^4-7a^2b^2 + kb^4 completely.

k = -18 (fyi)

Btw, I need to learn latex :P
Title: Re: cherylim23's methods question thread :)
Post by: brightsky on April 26, 2010, 09:59:02 pm: $a^4 - 7a^2b^2 + kb^4$

$= a^4 - 9a^2b^2 + 2a^2b^2 + kb^4$

$= a^2(a^2 - 9b^2) + b^2(2a^2 + kb^2)$

$= a^2(a + 3b)(a - 3b) + b^2 (2a^2 + kb^2)$

In order to factor this, we must try to make $b^2(2a^2+kb^2)$ have factors $(a+3b)$ and/or $(a-3b)$

$= a^2(a+3b)(a-3b) + 2b^2\left(a^2 + \frac{k}{2}b^2\right)$

$= a^2(a+3b)(a-3b) + 2b^2 \left(a^2 - \left(-\frac{k}{2}b^2\right)\right)$

Let $u^2 = -\frac{k}{2}$

$= a^2 (a+3b)(a-3b) + 2b^2 (a^2 - u^2b^2)$

$= a^2(a+3b)(a-3b) + 2b^2(a - ub)(a+ub)$

Now, it is apparent that in order to factorise THIS, we need $u = 3$

Because $u^2 = -\frac{k}{2}$

We need $3^2 = -\frac{k}{2}$

$9 = -\frac{k}{2}$

$k = -18$
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on April 26, 2010, 10:01:58 pm: Oh okay thanks! :D
I see, but aren't we supposed to use factor theorem for this question...?
Title: Re: cherylim23's methods question thread :)
Post by: brightsky on April 26, 2010, 10:13:45 pm: Oh! Yes! Much easier that way! ;)

Substitute in $3b$ into the expression for all the $a$ . Because it is a factor, by the factor theorem, it would equal to 0.

$(3b)^4 - 6(3b)^2(b^2) + kb^4 = 0$

$81b^4 - 54b^4 + kb^4 = 0$

$18b^4 + kb^4 = 0$

$k = -18$

:D
Title: Re: cherylim23's methods question thread :)
Post by: Yitzi_K on April 26, 2010, 10:32:56 pm: The factor theorem states that if (x-a) is a factor of f(x), then f(a)=0. So to use the factor theorem you just sub in the a value.
Title: Re: cherylim23's methods question thread :)
Post by: selim31 on April 29, 2010, 06:31:32 pm: @cherylim23
Hey, how did you go on your SAC? I had mine today and it was totally easy :)
Title: Re: cherylim23's methods question thread :)
Post by: 99.95 on May 02, 2010, 06:59:47 pm: how do i do this question:

Let P and Q be points on tge curve y=x^2 at which x=1 and x= 1+ h respectively. Express the gradient of the line PQ in terms of h and hence find the gradient of the tangent to the curve y=x^2 at x=1
Title: Re: cherylim23's methods question thread :)
Post by: Yitzi_K on May 02, 2010, 07:08:29 pm: Quote from: 99.95 on May 02, 2010, 06:59:47 pm
how do i do this question:

Let P and Q be points on tge curve y=x^2 at which x=1 and x= 1+ h respectively. Express the gradient of the line PQ in terms of h and hence find the gradient of the tangent to the curve y=x^2 at x=1

Gradient = y₂ - y₁ over x₂ - x₁.

This gives you (f(x+h) - f(x))/h

Simplify, (down to 2x+h) then let h=0 (even though h can't equal 0, we let it tend towards zero, which comes to the same thing) then sub in x=1.

This is called the First Principle method.
Title: Re: cherylim23's methods question thread :)
Post by: brightsky on May 02, 2010, 07:09:50 pm: Ok, the point P is $(1,y_1)$ for a y-coordinate $y_1$ and point Q is $(1+h, y_2)$ for a y-coordinate $y_2$ .

The points are on the curve $y = x^2$ hence it satisfies the equation.

Substitute in the coordinates:

$y_1 = 1^2 = 1$

$y_2 = (1+h)^2 = 1 + 2h + h^2$

Hence the points are P $(1,1)$ and Q $(1+h, 1+2h+h^2)$ .

To find the gradient of the line, use the formula: $\frac{y_2 - y_1}{x_2 - x_1}$ .

So Gradient = $\frac{(1+2h+h^2) - 1}{(1+h) - 1} = \frac{2h+h^2}{h} = \boxed{2+h}$

As for your second question, fastest way is the differentiate $y = x^2$

$\frac{dy}{dx} = 2x$

Substitute x = 1,

Gradient = $2 \times 1 = 2$
Title: Re: cherylim23's methods question thread :)
Post by: 99.95 on May 02, 2010, 07:12:29 pm: Quote from: Yitzi_K on May 02, 2010, 07:08:29 pm
Quote from: 99.95 on May 02, 2010, 06:59:47 pm
how do i do this question:

Let P and Q be points on tge curve y=x^2 at which x=1 and x= 1+ h respectively. Express the gradient of the line PQ in terms of h and hence find the gradient of the tangent to the curve y=x^2 at x=1

thanks for the help

Gradient = y₂ - y₁ over x₂ - x₁.

This gives you (f(x+h) - f(x))/h

Simplify, (down to 2x+h) then let h=0 (even though h can't equal 0, we let it tend towards zero, which comes to the same thing) then sub in x=1.

This is called the First Principle method.
Quote from: brightsky on May 02, 2010, 07:09:50 pm
Ok, the point P is $(1,y_1)$ for a y-coordinate $y_1$ and point Q is $(1+h, y_2)$ for a y-coordinate $y_2$ .

The points are on the curve $y = x^2$ hence it satisfies the equation.

Substitute in the coordinates:

$y_1 = 1^2 = 1$

$y_2 = (1+h)^2 = 1 + 2h + h^2$

Hence the points are P $(1,1)$ and Q $(1+h, 1+2h+h^2)$ .

To find the gradient of the line, use the formula: $\frac{y_2 - y_1}{x_2 - x_1}$ .

So Gradient = $\frac{(1+2h+h^2) - 1}{(1+h) - 1} = \frac{2h+h^2}{h} = \boxed{2+h}$

As for your second question, fastest way is the differentiate $y = x^2$

$\frac{dy}{dx} = 2x$

Substitute x = 1,

Gradient = $2 \times 1 = 2$

thanks for the help
Title: Re: cherylim23's methods question thread :)
Post by: brightsky on May 02, 2010, 07:14:33 pm: Yitzi_K's method is essentially the same. In full: Differentiation by first principles. Differentation comes from that. The derivative is essentially the gradient.
Title: Re: cherylim23's methods question thread :)
Post by: m@tty on May 02, 2010, 07:44:52 pm: Quote from: brightsky on May 02, 2010, 07:09:50 pm
As for your second question, fastest way is the differentiate $y = x^2$

$\frac{dy}{dx} = 2x$

Substitute x = 1,

Gradient = $2 \times 1 = 2$

The question says "hence", so you must use your answer to the previous part of the question.

You must let h=0 so $\frac{dy}{dx}=2+0=2$ .

You let h=0 because for the chord PQ to be a tangent it must have only one point of intersection with the parabola, ie. P=Q or h=0.

But you can certainly use brightsky's method to check your answer.
Title: Re: cherylim23's methods question thread :)
Post by: 99.95 on May 02, 2010, 09:20:10 pm: how do i differentiate:

y=(e^x^2)^5?
Title: Re: cherylim23's methods question thread :)
Post by: superflya on May 02, 2010, 09:25:26 pm: same as e^5x^2. now differentiate normally. it should be 10xe^5x^2
Title: Re: cherylim23's methods question thread :)
Post by: brightsky on May 02, 2010, 09:32:02 pm: $(e^{x^2})^5 = e^{5x^2}$

Let $u = 5x^2$

$\frac{d}{du} e^u = e^u \cdot u'$

$\implies e^{5x^2} \left(\frac{d}{dx}5x^2\right) = e^{5x^2} \cdot 10x$

$\implies 10e^{5x^2} x$

Unless you mean... $\left((e^x)^2\right)^5$ ...
Title: Re: cherylim23's methods question thread :)
Post by: m@tty on May 02, 2010, 09:45:22 pm: $y=\left(e^{x^2}\right)^5=e^{5x^2}$

So use chain rule with $u=5x^2$ $\frac{du}{dx}=10x$

$y=e^u$ $\frac{dy}{du}=e^u$

$\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx}=e^u \cdot 10 x$

$=e^{5x^2} \cdot 10x$

Alternatively you could use the chain rule twice,

$y=u^5$ where $u=e^{x^2}$ then $u=e^v$ where $v=x^2$ .

$\frac{dy}{du}=5u^4$ ; $\frac{du}{dv}=e^v$ ; $\frac{dv}{dx}=2x$

$\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$

$=5u^4 \cdot e^v \cdot 2x$

Substitute back in:

$=5\left(e^{x^2}\right)^4 \cdot e^{x^2} \cdot 2x$

$=2x \cdot \left(e^{x^2}\right)^5=10x \cdot e^{5x^2}$
Title: Re: cherylim23's methods question thread :)
Post by: 99.95 on May 06, 2010, 01:40:38 pm: how do i differentiate:

y=e^xsin^2x
Title: Re: cherylim23's methods question thread :)
Post by: the.watchman on May 06, 2010, 01:46:21 pm: Quote from: 99.95 on May 06, 2010, 01:40:38 pm
how do i differentiate:

y=e^xsin^2x

$\frac{dy}{dx} = e^x \cdot \frac{d}{dx}(\sin ^2 x) + \sin ^2 x \cdot \frac{d}{dx}(e^x)$

$= e^x \cdot 2\sin x \cdot \frac{d}{dx}(\sin x) + \sin ^2 x \cdot e^x$

$= e^x \cdot 2\sin x \cdot \cos x + \sin ^2 x \cdot e^x$

$= e^x \cdot \sin x (2\cos x + \sin x)$

This is a combination of the product and chain rules
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on May 09, 2010, 01:01:36 am: A few questions :)

What is the relation type of x = y^2?

8. State the range of the relation x2 + y2 > 9 when x, y is an element of R.

A.   [0, infinity)
B.   (– infinity, 0)
C.   (– infinity, 0]
D.   R
E.   (– infinity, –3) union (3, infinity)

7. The graph of a function with rule y = f(x) has one asymptote with equation x=4. The graph of the inverse function will have:
A no asymptote
B a vertical asymptote with equation x = 4
C a horizontal asymptote with equation y = 4
D a vertical asymptote with equation x = 1/4
E a horizontal asymptote with equation x = ¼

13. State the maximion with rule f (x) = 4 / ( rt (x-7) )
A R\{7
B R\{7}
C R+
D [7, infinity)
E (7, infinity)

Thanks, I'm pretty sleep now so I can't solve some of them :P
Title: Re: cherylim23's methods question thread :)
Post by: m@tty on May 09, 2010, 01:10:23 am: Well $x=y^2 \implies y=\pm\sqrt{x}$

So it's relation can be described as a square root in x.

Go look at the.watchman's post..

$x^2+y^2>9$ is a region involving all points outside the circle of radius 3 centered about the origin, so there are points below, above, as well as on either side of this circle. There are points of all y-values. So the range is $\mathbb{R}$ .

The inverse involves a reflection in the line y=x, swapping x and y. So an asymptote at $x=4$ becomes a horizontal asymptote at $y=4$ .

And what is the last question?? ...

I'm guessing it is the maximal domain of f...

Well, $f(x)=\frac{4}{\srqt{x-7}}$

Firstly for $\sqrt{x-7}$ to be defined $x-7\geq 0$ . But it is also the denominator, which cannot equal zero, therefore it cannot be zero. So the relevant domain is $x-7>0$ . So $x>7$ , or $x\in (7,\infty)$ .
Title: Re: cherylim23's methods question thread :)
Post by: the.watchman on May 09, 2010, 08:55:04 am: Hey m@tty, I thought the first answer might be a "one-to-many" relation, that's how I interpreted the question :)

Also:
Quote from: m@tty on May 09, 2010, 01:10:23 am
So $x>7$ , or $x\in (0,\infty)$ .

You must have been a bit sleepy - $x \in (7,\infty)$ ? :P
Title: Re: cherylim23's methods question thread :)
Post by: m@tty on May 09, 2010, 02:06:30 pm: Quote from: the.watchman on May 09, 2010, 08:55:04 am
Hey m@tty, I thought the first answer might be a "one-to-many" relation, that's how I interpreted the question :)

Also:
Quote from: m@tty on May 09, 2010, 01:10:23 am
So $x>7$ , or $x\in (0,\infty)$ .

You must have been a bit sleepy - $x \in (7,\infty)$ ? :P

Haha, yep..
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on May 09, 2010, 03:10:58 pm: thanks guys :D

And the.watchman's answer was what it was supposed to be for the type of relation qn xP

I'm a little confused with this question and the one I gave above.
The maximal domain for f(x)= rt (x+1) is best given by:
A) R+ union 0
B) R+
C) (-infinity, -1]
D) [-1, infinity)
E) [1, infinity)

The answer is D but I suppose it's wrong...?
Title: Re: cherylim23's methods question thread :)
Post by: brightsky on May 09, 2010, 03:18:31 pm: It's right.

The maximal domain is the largest range of x. Since 1 + x is under a square root sign, $1 + x \ge 0$ , so $x \ge -1$ . This translates to interval notations as [-1,infty).
Title: Re: cherylim23's methods question thread :)
Post by: brightsky on May 09, 2010, 03:19:25 pm: Quote from: cherylim23 on May 09, 2010, 03:17:03 pm
Quote from: cherylim23 on May 09, 2010, 01:01:36 am
7. The graph of a function with rule y = f(x) has one asymptote with equation x=4. The graph of the inverse function will have:

btw, know any graphs which have only one asymptote...?

$y = e^x$
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on May 09, 2010, 03:31:18 pm: Quote from: cherylim23 on May 09, 2010, 01:01:36 am
8. State the range of the relation x2 + y2 > 9 when x, y is an element of R.

A. [0, infinity)
B. (– infinity, 0)
C. (– infinity, 0]
D. R
E. (– infinity, –3) union (3, infinity)

Hmm, I still don't get why its not E.
Title: Re: cherylim23's methods question thread :)
Post by: brightsky on May 09, 2010, 03:41:03 pm: Technically the answer is D, isn't it? It's already stated that $x,y \in \mathbb{R}$ .

$x^2 + y^2 > 9 \implies y^2 > 9 - x^2$

Note the greater than sign. Hence y can be any number.
Title: Re: cherylim23's methods question thread :)
Post by: m@tty on May 09, 2010, 04:33:34 pm: Think about it graphically, the relation $x^2+y^2>9$ denotes all points whose distance from the origin is greater than 3. As $x^2+y^2>9 \implies \sqrt{x^2+y^2}>3$ . Now, $\sqrt{x^2+y^2}$ can be recognised as the hypotonuse of a right angled triangle. So the relation includes any point where a line from the origin is greater than 3 units in length.

As seen in the attached image, points of all values of y satisfy this requirement. Therefore the range is $\mathbb{R}$ .
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on May 09, 2010, 04:52:46 pm: I partially understood the front part.
But if you let x and y = 2,

2^2 + 2^2 = 8
8 < 9 and 8 is not > 9.

So the values between (2, -2) shouldn't be included?
Title: Re: cherylim23's methods question thread :)
Post by: 99.95 on May 09, 2010, 04:57:18 pm: how do u get

x>0 and x< -2 fo this question

let f(x) = x^3 + 3x^2 - 1

find: {x:f'(x) > 0}
Title: Re: cherylim23's methods question thread :)
Post by: m@tty on May 09, 2010, 04:57:41 pm: That point is not included, for the reason you stated.

But consider the point $(2,4)$

$2^2+4^2=20 > 9$ and hence it is included.

Similarly, the point $(4,0)$ is also included. In fact, any point outside of the circle is included, even those to the left and right. Hence there are points involved in the region for all real values of y.
Title: Re: cherylim23's methods question thread :)
Post by: m@tty on May 09, 2010, 05:00:26 pm: Quote from: 99.95 on May 09, 2010, 04:57:18 pm
how do u get

x>0 and x< -2 fo this question

let f(x) = x^3 + 3x^2 - 1

find: {x:f'(x) > 0}

$f'(x)=3x^2+6x$

$f'(x)>0$

$3x(x+2)>0$

So the intercepts are at $0$ and $-2$ , and it is an upright parabola.

Hence it is greater then zero before the lesser root and above the greater root. ie. $x<-2 \cup x>0$

You can sketch to see the result more visually.
Title: Re: cherylim23's methods question thread :)
Post by: 99.95 on May 09, 2010, 05:06:53 pm: thanks m@tty

how do i do this question:

y= X / 1-x

write dy/dx in terms of y
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on May 09, 2010, 05:14:18 pm: Oh I think I get it :) thanks :D
Basically, they are just asking the y values that are outside the circle for > 9?

However, for option E, since -3 and 3 are not included, they could be counted as outside the circle too?
Or if you don't mind explaining why option E is not right:P
Title: Re: cherylim23's methods question thread :)
Post by: brightsky on May 09, 2010, 05:50:19 pm: Quote from: 99.95 on May 09, 2010, 05:06:53 pm
thanks m@tty

how do i do this question:

y= x / 1-x

write dy/dx in terms of y

$y = \frac{x}{1-x} = \frac{1}{x-1} + 1$ ....(1)

$\frac{dy}{dx} = -\frac{1}{(x-1)^2} + 0 = -\frac{1}{(x-1)^2} = -\frac{1}{x-1} \cdot \frac{1}{x-1}$ ...(2)

From (1):

$y - 1 = \frac{1}{x-1}$

Sub that into (2):

$\frac{dy}{dx} = -(y - 1) \cdot (y - 1) = -(y-1)^2$

Hope this is right..
Title: Re: cherylim23's methods question thread :)
Post by: /0 on May 09, 2010, 08:11:19 pm: $y = \frac{x+(1-x)-(1-x)}{1-x} = \frac{1}{1-x}-1=-\frac{1}{x-1}-1$

Apart from the sign, the rest seems alright
Title: Re: cherylim23's methods question thread :)
Post by: 99.95 on May 09, 2010, 08:26:46 pm: correct me if i'm wrong:

for linear approximations:

you must always choose the closet value u can square root. u CANNOT choose any number. E.g. square root 120. can't choose 100 and 20. must be 121 - 1
Title: Re: cherylim23's methods question thread :)
Post by: Mao on May 09, 2010, 08:33:23 pm: You can choose 100, but your answer won't be accurate.

Generally, choose a pair of values (x,f(x)) that you accurately know to answer to.
Title: Re: cherylim23's methods question thread :)
Post by: m@tty on May 09, 2010, 09:17:30 pm: Quote from: cherylim23 on May 09, 2010, 05:14:18 pm
Oh I think I get it :) thanks :D
Basically, they are just asking the y values that are outside the circle for > 9?

However, for option E, since -3 and 3 are not included, they could be counted as outside the circle too?
Or if you don't mind explaining why option E is not right:P

Well, what about the point $(3,3)$ ?? Here the sum of squares is 18, which is greater than nine and hence included in the region. Similarly $(3,-3)$ and $(10,2)$ and $(-50,-1)$ are all included in this region.

Every single point on the plane that lies outside the circle is included in this region. That's because this circle is the exact line for which the sum of squares is 9, or the distance from the origin is three. Therefore every point outside of this circle satisfies the original requirement. See attached diagram.
Title: Re: cherylim23's methods question thread :)
Post by: 99.95 on May 09, 2010, 09:19:46 pm: is this correct:

for approximate change and approximate increase: use hf'(x)

for approximate value: use f(x) + h'f(x)
Title: Re: cherylim23's methods question thread :)
Post by: TrueTears on May 09, 2010, 09:20:54 pm: Quote from: m@tty on May 09, 2010, 04:33:34 pm
Think about it graphically, the relation $x^2+y^2>9$ denotes all points whose distance from the origin is greater than 3. As $x^2+y^2>9 \implies \sqrt{x^2+y^2}>3$ . Now, $\sqrt{x^2+y^2}$ can be recognised as the hypotonuse of a right angled triangle. So the relation includes any point where a line from the origin is greater than 3 units in length.

As seen in the attached image, points of all values of y satisfy this requirement. Therefore the range is $\mathbb{R}$ .
i like the implicit generator that /0 provided too, so slick
Title: Re: cherylim23's methods question thread :)
Post by: m@tty on May 09, 2010, 09:24:12 pm: Quote from: 99.95 on May 09, 2010, 09:19:46 pm
is this correct:

for approximate change and approximate increase: use hf'(x)

for approximate value: use f(x) + h'f(x)

Yeah. From memory:

$f(x+h) \approx f(x)+hf'(x)$

So the known value is $f(x)$ and the approximate change is $hf'(x)$ .
Quote from: TrueTears on May 09, 2010, 09:20:54 pm
Quote from: m@tty on May 09, 2010, 04:33:34 pm
Think about it graphically, the relation $x^2+y^2>9$ denotes all points whose distance from the origin is greater than 3. As $x^2+y^2>9 \implies \sqrt{x^2+y^2}>3$ . Now, $\sqrt{x^2+y^2}$ can be recognised as the hypotonuse of a right angled triangle. So the relation includes any point where a line from the origin is greater than 3 units in length.

As seen in the attached image, points of all values of y satisfy this requirement. Therefore the range is $\mathbb{R}$ .
i like the implicit generator that /0 provided too, so slick

Indeed, it is very good :P
Title: Re: cherylim23's methods question thread :)
Post by: /0 on May 10, 2010, 10:35:35 am: Huh what implicit generator? o_O
Title: Re: cherylim23's methods question thread :)
Post by: TrueTears on May 10, 2010, 11:12:44 am: implicit grapher man, pr0 pr0 pr0 in the mathematics forum
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on June 04, 2010, 07:31:36 pm: The relation: $x^2 + (y+3)^2 = 9$ , can be changed to a function by restricting the:
A domain to [-3, 0]
B domain to [-6, -3]
C range to [-3, 3]
D range to [-3, 0]
E domain to [-3, 0] and range to [-3, 3]

I think there's a problem with this questions, but anyway, the answer is apparently D.

This is a really simple question.
The chalk box of a teacher contains 8 white sticks of chalk, 3 yellow sticks and 2 blue sticks. If each stick has an equal chance of being selected, the probability the teacher randomly selects a yellow stick is:
A 0.2
B 0.25
C 0.3
D 0.33
E 0.4

My answer is $3/13$ which I'm positively sure it's right. Correct answer is A eh? D:

~~Edit: How do get rid of the HTML code btw? ;)~~
Title: Re: cherylim23's methods question thread :)
Post by: m@tty on June 04, 2010, 07:36:02 pm: $x^2+(y+3)^2=3$ is a circle of radius 3 and center $(0,-3)$ .

A function means that there is only one y-value for all x-values. If you split the circle at $y=-3$ you are left with two functions. Each of domain $[-3,3]$ and range of either $[-3,0]$ or $[-6,-3]$ . Therefore D is correct.
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on June 04, 2010, 07:37:46 pm: I thought C, D and E are all possible answers.

Edit: Oops, definitely not E. Because it'll be a one-to-many relation.
Looking at this thread, I seem to ask a lot of circle questions hahah
Title: Re: cherylim23's methods question thread :)
Post by: m@tty on June 04, 2010, 07:39:25 pm: I'm not sure if restricting the range to $-3,3$ is even feasible, lol. The original range is $[-6,0]$ ...
Title: Re: cherylim23's methods question thread :)
Post by: m@tty on June 04, 2010, 07:41:35 pm: For the numbers you have given the Pr is indeed 3/13. If the answer is meant to be A I assume that there was meant to two more sticks of chalk between white and blue.
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on October 22, 2010, 12:02:38 am: Some questions~~

16. The equivalent in degrees of 0.25 radians is
A. 14.32 B. 57.29 C. 90 D. 180 E. none of these

Answer is A but shouldn't it be $\frac{\pi}{4}$ ? -confused-

In Victoria, one of two daily morning newspapers claims that 85% of households that subscribe to the newspaper have their papers delivered each morning before 7.30am. In a random sample of 50 households that subscribe to this daily newspaper, the probability that exactly 45 receive their paper before 7.30am is given by
A. $0.85^{45} \times 0.15^5$
B. $0.85 \times {50 \choose 45} \times 0.15^5$
C. ${50 \choose 45} \times 0.85^5 \times 0.15^{45}$
D. ${50 \choose 45} \times 0.85^{45} \times 0.15^5$
E. $\frac{{85 \choose 45}{15 \choose 5}}{{100 \choose 50}}$

22. The probability of a soccer team scoring a goal from a penalty shot is 0.78.
During a particular game the team is awarded four penalty shots. The probability that they score three goals from these penalty shots is closest to:
A. 0.1044
B. 0.3702
C. 0.4176
D. 0.4746
E. 0.7878

All help is appreciated! :)
Title: Re: cherylim23's methods question thread :)
Post by: 98.40_for_sure on October 22, 2010, 12:17:00 am: For your 1st question:
Conversion from radians to degrees is x180 then /pi
Thus (0.25 x 180)/pi = 14.32

For your 2nd question:
Let X be the number of households that receive their paper before 7:30am
X~Bi(n=50,p=0.85)
So by formula
Pr(X=45) = nCr(50,45) x (0.85)^45 x (0.15)^5
Thus, C

For your 3rd question:
Let Y be the number of goals scored from penalty shots
Y~Bi(n=4,p=0.78)
Pr(Y=3) = nCr(4,3) x (0.78)^3 x (0.22)^1
= 0.4176
Thus, C
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on October 22, 2010, 12:19:50 am: Thanks!

However, I don't get your notation for: X~Bi(n=50,p=0.85) and Y~Bi(n=4,p=0.78).

And is 0.25 radian = $\frac{\pi}{4}$ ?
Title: Re: cherylim23's methods question thread :)
Post by: Blakhitman on October 22, 2010, 12:21:24 am: Quote from: cherylim23 on October 22, 2010, 12:02:38 am
Some questions~~

16. The equivalent in degrees of 0.25 radians is
A. 14.32 B. 57.29 C. 90 D. 180 E. none of these

Answer is A but shouldn't it be $\frac{\pi}{4}$ ? -confused-

to convert to degrees multiply by $\frac{180}{\pi}$

In Victoria, one of two daily morning newspapers claims that 85% of households that subscribe to the newspaper have their papers delivered each morning before 7.30am. In a random sample of 50 households that subscribe to this daily newspaper, the probability that exactly 45 receive their paper before 7.30am is given by
A. $0.85^{45} \times 0.15^5$
B. $0.85 \times {50 \choose 45} \times 0.15^5$
C. ${50 \choose 45} \times 0.85^5 \times 0.15^{45}$
D. ${50 \choose 45} \times 0.85^{45} \times 0.15^5$
E. $\frac{{85 \choose 45}{15 \choose 5}}{{100 \choose 50}}$

X~Bi(50,0.85) and we know $pr(X=x)={n \choose x} \times p^{x} \times q^{n-x}$
so in this case $pr(X=45)={50 \choose 45} \times 0.85^{45} \times 0.15^5$ so D

22. The probability of a soccer team scoring a goal from a penalty shot is 0.78.
During a particular game the team is awarded four penalty shots. The probability that they score three goals from these penalty shots is closest to:
A. 0.1044
B. 0.3702
C. 0.4176
D. 0.4746
E. 0.7878

Again, X~Bi(4,0.78) so to find $pr(x=3)={4 \choose 3} \times 0.78^{3} \times 0.22^1$

All help is appreciated! :)
Title: Re: cherylim23's methods question thread :)
Post by: 98.40_for_sure on October 22, 2010, 12:22:08 am: Quote from: cherylim23 on October 22, 2010, 12:19:50 am
Thanks!

However, I don't get your notation for: X~Bi(n=50,p=0.85) and Y~Bi(n=4,p=0.78).

And is 0.25 radian = $\frac{\pi}{4}$ ?

Well it just says X is a binomial distribution, where the number of trials is 50, and the probability is 0.85.

No you're confusing a radian with pi. A radian is 180/pi. It's 50something degrees roughly
Title: Re: cherylim23's methods question thread :)
Post by: Blakhitman on October 22, 2010, 12:26:04 am: Nope, but if this was CAS active then you would naturally use binompdf on your CAS (for the last question in this case).
Title: Re: cherylim23's methods question thread :)
Post by: 98.40_for_sure on October 22, 2010, 12:27:00 am: You could always list out all the possible combinations :P
there are... 2118760 of them
Title: Re: cherylim23's methods question thread :)
Post by: TyErd on October 22, 2010, 09:49:15 am: how would you do this one.
Title: Re: cherylim23's methods question thread :)
Post by: fady_22 on October 22, 2010, 10:01:23 am: Pr(odd)=0.3+0.25=0.55

Let O~Bi(5,0.55)
Pr(O>=3)=Pr(O=3)+Pr(O=4)+P(O=5)
You should get E.
Title: Re: cherylim23's methods question thread :)
Post by: TyErd on October 22, 2010, 10:21:34 am: okay thanks very much
Title: Re: cherylim23's methods question thread :)
Post by: itolduso on October 23, 2010, 10:06:20 pm: yes, velocity at t=2 is undefined
Title: Re: cherylim23's methods question thread :)
Post by: brightsky on October 23, 2010, 10:07:26 pm: Quote from: itolduso on October 23, 2010, 10:06:20 pm
yes, velocity at t=2 is undefined
Yes, that's right. My bad.
Title: Re: cherylim23's methods question thread :)
Post by: TrueTears on October 24, 2010, 02:43:20 am: Let y = f(x) be the curve (assuming it's a function of a single variable)

$f'(x) \propto x \implies f'(x) = kx \ \text{for some constant k}$

The rest should be trivial right? Just integrate and then some substitutions.
Title: Re: cherylim23's methods question thread :)
Post by: brightsky on December 19, 2010, 10:49:35 pm: 2x + 9y = 3 ---> 2x = 3 - 9y ---> x = (3 - 9y)/2

Sub into curve, (3 - 9y)y/2 + y + 2 = 0 -->y = -4/9 or y = 1, sub back in you get x = 7/2 or x = -3

So your points P and Q are (7/2, -4/9) and (-3, 1)

Now just implicit diff: y + xy' + y' = 0, xy' + y' = -y, y'(x + 1) = -x'y, y' = -y/(x+1)

Then sub and your done.
Title: Re: cherylim23's methods question thread :)
Post by: Andiio on December 19, 2010, 11:45:46 pm: ..The gradient of a curve varies directly as x? o_o Does that just mean that m = x?
Title: Re: cherylim23's methods question thread :)
Post by: aznxD on December 20, 2010, 12:24:18 am: Quote from: Andiio on December 19, 2010, 11:45:46 pm
..The gradient of a curve varies directly as x? o_o Does that just mean that m = x?

No direct variation is where if $a \propto x \ then a=kx$ where k is a constant

So let the curve of the graph be f(x)
Therefore the gradient of the curve $f'(x) \propto x \ so \ f'(x) = kx \ \text{where k is a constant}$
Then as TT mentioned, you just integrate f'(x) and sub in the coordinate values to find the equation of the curve
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on January 10, 2011, 04:05:59 pm: I don't know if this makes any sense...
but we all know how linear graphs are always one-to-one functions, and quadratic graphs are always many-to-one functions for the domain R, however, cubic graphs may be one-to-one functions or many-to-one functions across the domain R as well.

So, for all cubic graphs with an inflection point, they are definitely one-to-one functions. Otherwise, vice versa. Is this true for all cases?
Title: Re: cherylim23's methods question thread :)
Post by: pi on January 10, 2011, 06:02:18 pm: Your understanding is correct. Cubic graphs can either be one-to-one (inflection form only) or many-to-one (general shape), and its a point a lot of people would overlook (or not care about).
Quote from: cherylim23 on January 10, 2011, 04:05:59 pm
I don't know if this makes any sense...

...

Is this true for all cases?

Don't really understand that last question. If you want to know if all cubics in the form f(x)=ax^3 (on Cartesian plane) are one-to-one, that is correct. This property is true for all f(x)-ax^n, where n is an odd natural number (they all have inflection points then).
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on January 10, 2011, 07:25:55 pm: Thanks Rohitpi. What I meant was that for all cubic graphs with an inflection point, they'd definitely be one-to-one functions yes?
I guess you pretty much answered it

Also, we know that all cubic graphs will at least have one solution, rational or irrational. Does this lead to: all cubic graphs with an inflection point would have one and only one solution?
Title: Re: cherylim23's methods question thread :)
Post by: pi on January 10, 2011, 07:38:10 pm: Quote from: cherylim23 on January 10, 2011, 07:25:55 pm
Thanks Rohitpi. What I meant was that for all cubic graphs with an inflection point, they'd definitely be one-to-one functions yes?
I guess you pretty much answered it

Also, we know that all cubic graphs will at least have one solution, rational or irrational. Does this lead to: all cubic graphs with an inflection point would have one and only one solution?

All cubic graphs have 3 solutions: one real and two complex/imaginary, two real and one complex/imaginary or all three real (dunno if all three can be complex/imaginary, but who cares?). A graph with an inflection point has 3 (real) solutions, but only one distinct solution. There is an important difference there.

Learnt that here:
Quote from: kyzoo
...

• $(x-a)(x-b)^2$ has 3 real solutions $(x = a, b, b)$ . But 2 distinct real solutions $(x = a, b)$

...
^Check out his tips, they are really good (~~MM resources thread and scroll down~~ go here)
Title: Re: cherylim23's methods question thread :)
Post by: pi on January 10, 2011, 07:55:27 pm: Quote from: cherylim23 on January 10, 2011, 07:47:54 pm
Haha, I was referring to one and only one REAL solution. Sorry if I didn't make myself clear enough.
And nope, it's not possible for a cubic graph to have three complex solutions

Thought three complex solutions would be weird (thanks for the clarification)... On the solution, yep, one real distinct solution for an inflection cubic, but remember that are really three real solutions (just one -one real one- is repeated three times) <--- wording is important. I never thought of that (the 'distinct stuff') until I read kyzoo's tips... Don't think its taught too well in textbooks
Title: Re: cherylim23's methods question thread :)
Post by: brightsky on January 10, 2011, 08:00:24 pm: Quote from: Rohitpi on January 10, 2011, 07:55:27 pm
Quote from: cherylim23 on January 10, 2011, 07:47:54 pm
Haha, I was referring to one and only one REAL solution. Sorry if I didn't make myself clear enough.
And nope, it's not possible for a cubic graph to have three complex solutions

Thought three complex solutions would be weird (thanks for the clarification)... On the solution, yep, one real distinct solution for an inflection cubic, but remember that are really three real solutions (just one -one real one- is repeated three times) <--- wording is important. I never thought of that (the 'distinct stuff') until I read kyzoo's tips... Don't think its taught too well in textbooks

It's actually referred to in Essentials, not sure about MQ. It's the fundamental theorem of algebra.
Title: Re: cherylim23's methods question thread :)
Post by: pi on January 10, 2011, 08:02:48 pm: Quote from: brightsky on January 10, 2011, 08:00:24 pm
It's actually referred to in Essentials, not sure about MQ. It's the fundamental theorem of algebra.

Haha, I stand corrected. Sorry cherylim23... (bad info there). And my school uses MQ...
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on January 10, 2011, 08:05:07 pm: Quote from: Rohitpi on January 10, 2011, 07:55:27 pm
Thought three complex solutions would be weird (thanks for the clarification)... On the solution, yep, one real distinct solution for an inflection cubic, but remember that are really three real solutions (just one -one real one- is repeated three times) <--- wording is important. I never thought of that (the 'distinct stuff') until I read kyzoo's tips... Don't think its taught too well in textbooks
Yes, think about it this way. When the discriminant is zero, there are actually two solutions, but two repeated solutions that's all.

Hence arises the confusion when the question states, for what values of (insert variable) would the equation have two solutions.
I don't know whether to let discriminant > 0 or discriminant >= 0
Title: Re: cherylim23's methods question thread :)
Post by: pi on January 10, 2011, 08:06:57 pm: Quote from: cherylim23 on January 10, 2011, 08:05:07 pm
Yes, think about it this way. When the discriminant is zero, there are actually two solutions, but two repeated solutions that's all.

Hence arises the confusion when the question states, for what values of (insert variable) would the equation have two solutions.
I don't know whether to let discriminant > 0 or discriminant >= 0

Assuming quadratics (as cubics also have discriminants, but big ones...), two solutions would be dis > 0.
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on January 10, 2011, 08:09:28 pm: Yeah, but you could also say that when Dis = 0, there are two real solutions. Just that solutions are the same...
Title: Re: cherylim23's methods question thread :)
Post by: iNerd on January 10, 2011, 08:10:16 pm: Quote from: cherylim23 on January 10, 2011, 08:05:07 pm
Quote from: Rohitpi on January 10, 2011, 07:55:27 pm
Thought three complex solutions would be weird (thanks for the clarification)... On the solution, yep, one real distinct solution for an inflection cubic, but remember that are really three real solutions (just one -one real one- is repeated three times) <--- wording is important. I never thought of that (the 'distinct stuff') until I read kyzoo's tips... Don't think its taught too well in textbooks
Yes, think about it this way. When the discriminant is zero, there are actually two solutions, but two repeated solutions that's all.

Hence arises the confusion when the question states, for what values of (insert variable) would the equation have two solutions.
I don't know whether to let discriminant > 0 or discriminant >= 0
IMO the question should state two distinct solutions...otherwise let discriminant > O
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on January 10, 2011, 08:36:36 pm: Also, for a cubic polynomial, when there are two real and equal roots, the last root must definitely be real and not complex yes? Is there any proof for this?
Title: Re: cherylim23's methods question thread :)
Post by: m@tty on January 10, 2011, 09:26:39 pm: If all co-efficients are real, then yes, that is correct.

Say the two real roots of a polynomial f with real co-efficients are at x=a, and a third root is x=b+ci.

$f(x)=(x-a)^2(x-(b+ci))$

$f(x)=(x^2-2ax+a^2)(x-(b+ci))$

$f(x)=x^3-2ax^2-(b+ci)x^2+2ax(b+ci)+a^2x-a^2(b+ci)$

But since the coefficients must all be real, c must equal 0.

So the third root is also "real" rather than "complex".

(I will remind you here that all real numbers are in fact complex numbers as $\mathbb{R} \subset \mathbb{C}$ . ie. 2=2+0i)

If however, there were imaginary coefficients, the third root could be "complex".

This is more like an algebraic proof - but there are much more intuitive ways of understanding it. Think of it graphically, or if you are familiar with the conjugate root theorem for polynomials with real coefficients, then it is clearly seen that there must be a third real root.
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on January 15, 2011, 11:35:56 am: ^ Thanks m@tty :)

As for the attached question, I don't get why the answer is E and not A. They are both technically right...
Title: Re: cherylim23's methods question thread :)
Post by: pi on January 15, 2011, 11:39:27 am: Quote from: cherylim23 on January 15, 2011, 11:35:56 am

As for the attached question, I don't get why the answer is E and not A. They are both technically right...

~~They are both right, especially as no horizontal scaling is given...~~ See m@tty's post, I made a stupid error...
Title: Re: cherylim23's methods question thread :)
Post by: m@tty on January 15, 2011, 11:54:45 am: Ah, no. It is asymptotic to y=0 - therefore only E.

If A was $2^{-x+1}$ , then it would be another possibility with the given info. But it is not.
Title: Re: cherylim23's methods question thread :)
Post by: iNerd on January 15, 2011, 11:57:55 am: Quote from: m@tty on January 15, 2011, 11:54:45 am
Ah, no. It is asymptotic to y=0 - therefore only E.

If A was $2^{-x+1}$ , then it would be another possibility with the given info. But it is not.
Haven't done exponentials in ages but does the +1 in A indicate that the asymptote is at y = 1 ?
Title: Re: cherylim23's methods question thread :)
Post by: pi on January 15, 2011, 11:59:06 am: Quote from: ATAR on January 15, 2011, 11:57:55 am
Quote from: m@tty on January 15, 2011, 11:54:45 am
Ah, no. It is asymptotic to y=0 - therefore only E.

If A was $2^{-x+1}$ , then it would be another possibility with the given info. But it is not.
Haven't done exponentials in ages but does the +1 in A indicate that the asymptote is at y = 1 ?

Yep, that right. lol, can't believe I got that wrong... (better wrong now than in an exam though). Thanks m@tty, that was a really stupid error on my part.
Title: Re: cherylim23's methods question thread :)
Post by: m@tty on January 15, 2011, 12:00:33 pm: Haha, it's ok. Maths is actually really easy to make errors in..

Quote from: ATAR on January 15, 2011, 11:57:55 am
Quote from: m@tty on January 15, 2011, 11:54:45 am
Ah, no. It is asymptotic to y=0 - therefore only E.

If A was $2^{-x+1}$ , then it would be another possibility with the given info. But it is not.
Haven't done exponentials in ages but does the +1 in A indicate that the asymptote is at y = 1 ?

Yep.

You don't need to remember it. Think, as x becomes very large( $x\to \infty$ ) $2^{-x} \to 0$ hence $y\to 1$ .
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on January 15, 2011, 12:01:36 pm: I should facepalm myself because it took me so long and I never figured that out. If it was A, the asymptote would be at y=1 I assume.

Thanks again, matty.

/Edit: and thanks to everyone else
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on January 17, 2011, 02:58:20 pm: For this particular question attached, there may be more than one approach to part c)

How would any of you guys do it?

I did it differently from the worked solutions...

/Edit: Also, comparing these two graphs: $f(x) = \frac{1}{(x-4)^2}$ and $g(x) = \frac{1}{(x-4)^4}$
What's the transformation from f(x) to g(x)?

Thanks for reading
Title: Re: cherylim23's methods question thread :)
Post by: pi on January 17, 2011, 03:08:07 pm: I would have found the amplitude (and hence minimum) and then subbed in t=9 to prove it was a minimum. Then (just to be sure), would have done the gradient table thing to show that it is in fact a minimum.
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on January 17, 2011, 03:13:05 pm: Thanks Rohitpi. I wouldn't have thought of the gradient table though. :P

This is what the worked solutions showed - I reckon it's a longer method

/Edit: Actually, technically speaking, during that point of the year, say that question came out on a SAC, I don't think a gradient table would be required because we haven't learnt to differentiate circular functions 8-)
Title: Re: cherylim23's methods question thread :)
Post by: pi on January 17, 2011, 03:15:21 pm: Quote from: cherylim23 on January 17, 2011, 03:13:05 pm
Thanks Rohitpi. I wouldn't have thought of the gradient table though. :P

This is what the worked solutions showed - I reckon it's a longer method

I think my (or your) method is the reverse of that... Only difference is that ours would probably save a couple of minutes. I think with a gradient table, both would be marked equally.
Title: Re: cherylim23's methods question thread :)
Post by: m@tty on January 17, 2011, 04:31:53 pm: Quote from: cherylim23 on January 17, 2011, 02:58:20 pm
/Edit: Also, comparing these two graphs: $f(x) = \frac{1}{(x-4)^2}$ and $g(x) = \frac{1}{(x-4)^4}$
What's the transformation from f(x) to g(x)?

Thanks for reading

$g(x)=\left[f(x)\right]^2$
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on January 17, 2011, 05:06:51 pm: ^ Is that a dilation or some sort?
Because the graphs: $y = \log_{10}(x-3)^2$ and $y = 2\log_{10}(x-3)$ are technically speaking, equivalent of each other, but the latter is dilated by a factor of 2 from the x-axis...
Title: Re: cherylim23's methods question thread :)
Post by: nacho on January 17, 2011, 05:44:35 pm: Are these recent questions from MM or essentials? And what chapter?
Title: Re: cherylim23's methods question thread :)
Post by: vea on January 17, 2011, 07:51:48 pm: Quote from: cherylim23 on January 17, 2011, 05:06:51 pm
^ Is that a dilation or some sort?
Because the graphs: $y = \log_{10}(x-3)^2$ and $y = 2\log_{10}(x-3)$ are technically speaking, equivalent of each other, but the latter is dilated by a factor of 2 from the x-axis...

They are algebraically the same but graphically different... Not so sure how to explain this.
Title: Re: cherylim23's methods question thread :)
Post by: brightsky on January 17, 2011, 07:55:18 pm: They are both the same graph.
Title: Re: cherylim23's methods question thread :)
Post by: evaever on January 17, 2011, 08:04:50 pm: Quote from: cherylim23 on January 17, 2011, 05:06:51 pm

Because the graphs: $y = \log_{10}(x-3)^2$ and $y = 2\log_{10}(x-3)$ are technically speaking, equivalent of each other, .....

They are different, the square one has two branches
Title: Re: cherylim23's methods question thread :)
Post by: dude on January 17, 2011, 08:06:54 pm: I think the first one is defined for x < 3, but in the domain (3,infinity) they are the same
Title: Re: cherylim23's methods question thread :)
Post by: Andiio on January 17, 2011, 08:07:33 pm: IMO It's an issue of the domains, but they are the same graph
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on January 17, 2011, 08:42:42 pm: That wasn't what I meant. :P

Okay, maybe that example wasn't good enough.

But you would say that $y=\log_{10}(x-3)^{-1}$ and $y=-\log_{10}(x-3)$ are identical to each other algebraically and graphically.
Since the latter is reflected in the x-axis, would you say that the former is reflected in the x-axis too (although it doesn't directly imply so)?
Title: Re: cherylim23's methods question thread :)
Post by: dude on January 17, 2011, 08:45:24 pm: Is reflected in the x-axis the same as across the x-axis?
Title: Re: cherylim23's methods question thread :)
Post by: schnappy on January 17, 2011, 08:57:49 pm: $y = \log_{10}(x-3)^2$

$y = 2\log_{10}(x-3)$

As said they are the same, however the first is defied for x<3 because you sqaure a -ve you get a +ve, so you can take the log of that number. You should get the same graph, but also the same graph vertically reflected along the asymptote.
Title: Re: cherylim23's methods question thread :)
Post by: evaever on January 17, 2011, 09:14:19 pm: y=-logx is the reflection of y=logx in the x-axis, true

y=logx^-1 is the same function as y=-logx, so it is the reflection of y=logx in the x-axis, also true
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on January 17, 2011, 09:28:27 pm: ^Thanks evaever :)

Quote from: schnappy on January 17, 2011, 08:57:49 pm
$y = \log_{10}(x-3)^2$

$y = 2\log_{10}(x-3)$

As said they are the same, however the first is defied for x<3 because you sqaure a -ve you get a +ve, so you can take the log of that number. You should get the same graph, but also the same graph vertically reflected along the asymptote.

Does this imply that $y = \log_{10}(x-3)^2$ is dilated by a factor of 2 in the x-axis and also reflected in the y-axis or both or neither?
Title: Re: cherylim23's methods question thread :)
Post by: brightsky on January 17, 2011, 09:33:39 pm: Both. (Not y-axis but yeah. :p)
Title: Re: cherylim23's methods question thread :)
Post by: evaever on January 17, 2011, 11:30:43 pm: Does this imply that $y = \log_{10}(x-3)^2$ is dilated by a factor of 2 in the x-axis and also reflected in the y-axis or both or neither?
[/quote]

No, because $y = \log_{10}(x-3)^2$ is not the same function as $y = 2\log_{10}(x-3)$

Keep in mind you are comparing with $y = \log_{10}(x-3)$
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on January 27, 2011, 11:17:41 am: How would you solve this question?

$2^x = x+1$

Apparently, taking log doesn't work, but by inspection, I could tell x = 0 or 1
(Inspection is dodgy method IMO!)
Title: Re: cherylim23's methods question thread :)
Post by: pi on January 27, 2011, 11:47:16 am: Quote from: cherylim23 on January 27, 2011, 11:17:41 am
How would you solve this question?

$2^x = x+1$

Apparently, taking log doesn't work, but by inspection, I could tell x = 0 or 1
(Inspection is dodgy method IMO!)

I have no idea how to solve this either...

I did it graphically by graphing $y = 2^x -x -1$ (addition/subtraction of ordinate method) and the x-ints were 0 and 1 (your solutions)... But algebraically, I have no idea. I am interested in any algebriac solution too!
Title: Re: cherylim23's methods question thread :)
Post by: brightsky on January 27, 2011, 04:55:07 pm: I'm pretty sure it's a transcendental function. There are ways to solve it, namely Newton's method, but it's pretty complicated and can only get you approximations (although good approximations).
Title: Re: cherylim23's methods question thread :)
Post by: Water on January 27, 2011, 05:52:08 pm: How about rooting both sides by x? though the answer was only 1 ):
Title: Re: cherylim23's methods question thread :)
Post by: luken93 on January 28, 2011, 08:15:40 pm: yeah, don't think there is an easy solution:

http://www.wolframalpha.com/input/?i=2^x+%3D+x%2B1
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on February 20, 2011, 08:18:22 pm: Just a confirmation,

Dilated by a factor of n value from x-axis = Dilated by a factor of n value parallel to y-axis = Dilated by a factor of n value in the y-direction = Dilated by a factor of n value about the x-axis?

Hope that didn't seem too confusing!
Title: Re: cherylim23's methods question thread :)
Post by: Greatness on February 20, 2011, 08:23:36 pm: This 'Dilated by a factor of n value from x-axis = Dilated by a factor of n value parallel to y-axis' is true, i havent seen the other 2 tho. Also my teacher said that we dont need to know dilated by a factor of n parallel to x/y axis.
Title: Re: cherylim23's methods question thread :)
Post by: vea on February 20, 2011, 08:41:04 pm: Yes, they are equal. Though I should point out that the first and last one are the same o_O
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on February 21, 2011, 07:45:49 pm: Quote from: vea on February 20, 2011, 08:41:04 pm
Yes, they are equal. Though I should point out that the first and last one are the same o_O

oh, sorry! Thanks for pointing out, I meant to say, is "Dilated by a factor of n from x-axis = Dilated by a factor of n about x-axis"? :uglystupid2:
Title: Re: cherylim23's methods question thread :)
Post by: kamil9876 on February 21, 2011, 08:02:57 pm: Quote from: swarley on February 20, 2011, 08:23:36 pm
This 'Dilated by a factor of n value from x-axis = Dilated by a factor of n value parallel to y-axis' is true, i havent seen the other 2 tho. Also my teacher said that we dont need to know dilated by a factor of n parallel to x/y axis.

In my opinion I think parallel to y axis is worse because it doesn't specify where the dilation is from (though in most cases it is from the x-axis), for example if I said what do you get if you dilate the graph from of $y^2+x^2=1$ by a factor of 2 from the line $y=1$ then the dilation is parralel to the y-axis but it is not from the x-axis. :P
Title: Re: cherylim23's methods question thread :)
Post by: |ll|lll| on May 29, 2011, 05:22:33 pm: Stuck on a question!

Given that the expression $x^2 -5x +7$ leaves the same remainder whether divided by x-b or x-c where b =/= c, show that b + c = 5
Given further that 4bc = 21 and b > c, find the values of b and c.
Title: Re: cherylim23's methods question thread :)
Post by: b^3 on May 29, 2011, 05:56:11 pm: Ok so the same remainder is left implies that b²-5b+7=c²-5c+7 where c does not equal b
so b²-c²=5(b-c)
factorise LHS (b-c)*(b+c)=5(b-c)
now cancel the b-c
b+c=5

next part
4bc=21, b>c
so b=5-c
4(5-c)c=21
expand it 20c-4c²-21=0
so 4c²-20c+21=0
now factorise (2c-3)*(2c-7)=0
so c = 3/2 or c=7/2
therefore b= 7/2 or 3/2 but b>c
so b=7/2 and c =3/2