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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: TrueTears on December 03, 2008, 03:30:23 pm

Title: sketching graph + 2 functions questions
Post by: TrueTears on December 03, 2008, 03:30:23 pm: Q1. How would you sketch $\frac{1}{\sqrt{x}}$ ? Well, the question is actually find the range of $\frac{1}{\sqrt{x}}$ when $x\in R^+$ ? I was thinking about sketching the graph and work it out from there, but i have never met something like $\frac{1}{\sqrt{x}}$ . Is there a way to work out the range without sketching? If so, how would you sketch graphs like $\frac{n}{\sqrt{x}}$ where $n\in R$ in general ( without the use of a calc)?

Q2. Let a be a positive number, let $f:[2,\infty) --> R, f(x)= a-x$ and let $g:(-\infty,1] --> R, g(x)=x^2+a$ . Find all values of a for which $f \circ g$ and $g \circ f$ both exist.

I've gotten this far: $ran g \subset dom f$ ie, $ran g \subset [2,\infty)$ for $f \circ g$ to exist. And $ran f \subset dom g$ , ie $ran f \subset (-\infty,1]$ for $g \circ f$ to exist.

Also $dom f \circ g = (-\infty,1]$ and $dom g \circ f = [2,\infty)$ .

$f \circ g = f(g(x)) = a-x^2-a = -x^2$

$g \circ f = g(f(x)) = (a-x)^2+a = a^2-2ax+x^2+a$

How do you solve it from here? Or do you need to do it a completely different way?

Q3. Let $g:[b,2] --> R, g(x) = 1-x^2$ . If b is the smallest real value such that g has an inverse function, find b and $g^-^1(x)$

Many thanks!
Title: Re: sketching graph
Post by: bucket on December 03, 2008, 03:56:41 pm: I just sketch them, but I did methods cas.
Title: Re: sketching graph
Post by: dekoyl on December 03, 2008, 04:02:22 pm: Did you do GMA?
Sketch $\sqrt{x}$
The x-intercepts indicate an asymptote as $\frac{1}{0} = \infty$
So it looks like half a truncus (the right half of $\frac{1}{x^2}$ )
Title: Re: sketching graph
Post by: TrueTears on December 03, 2008, 04:07:36 pm: Quote from: dekoyl on December 03, 2008, 04:02:22 pm
Did you do GMA?
Sketch $\sqrt{x}$
The x-intercepts indicate an asymptote as $\frac{1}{0} = \infty$
So it looks like half a truncus (the right half of $\frac{1}{x^2}$ )

yeah i did GMA but we never learnt how to sketch 1/ sqrt (x) . i know how to sketch $\sqrt{x}$ Then how do u work out the asymptotes for 1\sqrt (x)
Title: Re: sketching graph + a composite function question
Post by: dekoyl on December 03, 2008, 05:49:23 pm: Quote from: TrueTears on December 03, 2008, 04:07:36 pm
yeah i did GMA but we never learnt how to sketch 1/ sqrt (x) . i know how to sketch $\sqrt{x}$ Then how do u work out the asymptotes for 1\sqrt (x)
It involves logic as well.
When $\sqrt{x} = 0$ , that is the x-intercept, right?
Now, at the x intercept, knowing $\sqrt{x} = 0$ , the graph of $\frac{1}{\sqrt{x}}$ would be $\frac{1}{0} = \infty$ . Because $\frac{1}{\sqrt{x}}$ never touches x = 0 (because you know that it'll be $\infty$ at x =0), it will be an asymptote there.

Try sketching on your calculator. It might help you understand what I'm trying to say more:
y1 = $\sqrt{x}$
y2 = $\frac{1}{\sqrt{x}}$

Oh shit I suck at explaining :( Hope you get it.
Title: Re: sketching graph + a composite function question
Post by: dekoyl on December 03, 2008, 05:52:29 pm: Also not really required but..

$\frac{1}{f(x)} = f(x)$ at $y = \pm1$
because
$\frac{1}{-1} = -1$
and
$\frac{1}{1} = 1$

So if you were to draw $\frac{1}{\sqrt{x}}$ and $\sqrt{x}$ on the same axis, they'll have a common point at $y = 1$
Title: Re: sketching graph + 2 functions questions
Post by: TrueTears on December 03, 2008, 06:35:26 pm: hmm yeah i kinda get that, like when i work out how to sketch hyperbolas or truncas what i do is: if i get something like
$\frac{1}{x-2} + 2$ the asymptotes will be at x=2 and y=2 since the graph is shifted across 2 to the right and 2 up, then because i also know the general shape of a hyperbola, i'd find the x and y intercepts and sketch it. And if it is just $\frac{1}{x}$ . The asymptotes are just the y axis and x axis ie, x=0 y=0. So if you get $\frac{1}{\sqrt{x}}$ what is the general shape of this family of graphs? Is it just the same as the 1st quadrant of a hyperbola except it gets steeper much faster?

So if you get \frac{1}{\sqrt{x+4}} -5. How would you work out the asymptotes here? Is it just x=-4 and y=-5. Then work out the y intercepts which = -4.5 and x intercept = -3.96 and sketch it with the general shape of a hyperbola except it gets steeper much faster?

Something like this?

(http://img266.imageshack.us/img266/9148/graphki7.png)
Title: Re: sketching graph + 2 functions questions
Post by: dekoyl on December 03, 2008, 06:42:58 pm: Quote from: TrueTears on December 03, 2008, 06:35:26 pm
So if you get \frac{1}{\sqrt{x+4}} -5. How would you work out the asymptotes here? Is it just x=-4 and y=-5. Then work out the y intercepts which = -4.5 and x intercept = -3.96 and sketch it with the general shape of a hyperbola except it gets steeper much faster?

Something like this?

(http://img266.imageshack.us/img266/9148/graphki7.png)

Yep that's right. :)
(But really, I don't think I've ever sketched it in methods.. or very rarely. I did this mostly in GMA)
Title: Re: sketching graph + 2 functions questions
Post by: kurrymuncher on December 03, 2008, 06:43:48 pm: Just think of it as a reciprocal. 1/sqr(x) is the same as sqr(X) flipped upside down.
Title: Re: sketching graph + 2 functions questions
Post by: /0 on December 03, 2008, 06:54:57 pm: Try graphing graphs of $y=\frac{1}{x^n},\ n>0$ for different values of $n$ . They all have the basic 'hyperbola' shape.

As $n$ gets bigger, the graph becomes steeper for $0<x<1$ and flatter for $x>1$ . The trend is kept for fractional powers, although if the index has an even number in the denominator, x cannot be negative, and if it has an even number in the numerator, y cannot be negative.

Q2

$ran f \subseteq (- \infty, 1]$ , $ran g \subseteq [2, \infty)$

So 1. we need to find $a$ so that any value of $f:[2 ,\infty) \rightarrow R, f(x)=a-x$ will only be a subset of $(-\infty, 1]$

The function $f(x)$ is a straight line and always decreasing, so the maximum value will be at its left endpoint, at $f(2) = a-2$ . Hence, we can say $ranf = (-\infty, a-2]$ . For it to have the correct range, we require $a-2 \leq 1 \Rightarrow a \leq 3$

and 2. we need to find $a$ so that any value of $g:(-\infty, 1] \rightarrow R, g(x) = x^2+a$ will only be a subset of $[2, \infty)$

For $g(x) = x^2+a$ , $a$ determines the vertical translation of the standard parabola with vertex at (0,0). We can say $rang = [a, \infty)$ . So to have the right range, $a \geq 2$ .

Hence the answer is $2 \leq a \leq 3$ .

Q3

For a function to have an inverse, it must be one-to-one. For a quadratic function, the graph is one-to-one up till the vertex, and after that it becomes many-to-one.

For $g(x) = 1-x^2$ , the vertex is at $x=0$ , hence the smallest real value of $b$ is $b=0$ .

For the inverse,

$x=1-y^2$

$y=\pm \sqrt{1-x}$

But as $rang^{-1} = domg = [0,2]$ , the sign has to be positive, so the inverse is

$g:[-3,1] \rightarrow R, g^{-1}(x) = \sqrt{1-x}$
Title: Re: sketching graph + 2 functions questions
Post by: TrueTears on December 03, 2008, 07:31:02 pm: ah yes thank you all :D

And one more Q

A cuboid tank is open at the top and the internal dimensions of its base are x and 2x m. The height is h m. The volume of the tank ix V cubic metres and the volume is fixed. Let $S m^2$ denote the internal surface area of the tank.

a) i. Find S in terms of x and h. $S=2x^2+6xh$
ii. Find S in terms of V and x. $V=2x^2h$ $h=\frac{V}{2x^2}$ so $S=2x^2+\frac{3V}{x}$

This is all fine, just the next few i can't get

b) State the maximal domain for the function defined by the rule in a) ii.how do you find the implied domain here?
c) if 2<x<15 find the maximum value of S if V= 1000 $m^3$
Title: Re: sketching graph + 2 functions questions
Post by: /0 on December 03, 2008, 07:49:58 pm: b)

$S = 2x^2+\frac{3V}{x}$

Hmmm the only real restriction I can see here is that $x>0$ (from a practical aspect lengths must be positive).
x cannot equal 0 but that case is already covered

c)

To find the maximum of a graph in a given domain you need to do 2 things:
1. Check the endpoints.
2. Check stationary points for local maximums.
If possible, look at the graph.

$S(2) = 1508$ , $S(15) = 650$

Differentiating with respect to x and solving equal to 0, stat. point is at $x=9.086$ , giving $S = 495.29$

So the maximum value of S is S = 1508. (although x cannot equal 2, the value x gives near 2 is so close to 1508 that the difference is negligible)
Title: Re: sketching graph + 2 functions questions
Post by: TrueTears on December 03, 2008, 07:53:02 pm: ah yes i was wondering why x = 2, because it said 2<x<15 so 2 is technically not included but yeah thanks

just another 3 questions :P

A1
$f(x)=\begin{cases} x^2-4 for x\in(-\infty,2) \\ x for x\in[2,\infty) \end{cases}$

if $h(x)= 2x$ find $f(h(x))$ . The answer is

$f(2x)=\begin{cases} 4x^2-4 for x<1 \\ 2x for x\ge1 \end{cases}$

where does the $x<1$ and $x\ge1$ come from?

Q2
Let $f: R \ \begin{Bmatrix} \frac{-d}{c}\end{Bmatrix} --> R, f(x)=\frac{ax+b}{cx+d}$ . What is the range of f(x) ?

Q3
Let $f(x)=\frac{px+q}{x+r}$ where $x \in R\setminus\begin{Bmatrix}-r,r\end{Bmatrix}$ . If $f(-x) = -f(x)$ for $x \neq 0$ , find the rule for $f(x)$ in terms of q
Title: Re: sketching graph + 2 functions questions
Post by: /0 on December 04, 2008, 03:21:45 pm: Q1

We require $ranh \subseteq domf$

i.e. $ranh \subseteq (-\infty , 2)$ for the first part and $ranh \subseteq [2, \infty)$ for the second part.

Now you have to find the domain from this range, (its a bit like finding the range in reverse)

For $h(x) = 2x$ , for the range $(-\infty, 2)$ , the domain must be restricted to $(-\infty , 1)$
and for the range $[2, \infty)$ , the domain must be restricted to $[1, \infty)$ . And that's how you get the domains for the hybrid function.

Q2.

If you perform long division on the fraction, you get $f(x) = \frac{bc-ad}{c(cx+d)}+\frac{a}{c}$

(http://img507.imageshack.us/img507/1176/longdivisionou6.jpg)

Since $a,b,c,d$ are all constants, this is just a hyperbola. The asymptotes are at $y = \frac{a}{c}$ and $x=\frac{-d}{c}$ .

So the range is $R \ \backslash \ \{\frac{a}{c}\}$ .

Q3.

I don't really know how to do this question.... this is as far as I got.

$f(-x) = -f(x) \Rightarrow \frac{-px+q}{-x+r} = -\frac{px+q}{x+r}$

$(-px+q)(x+r) = -(px+q)(-x+r)$

$-px^2+(q-pr)x+qr = px^2+(q-pr)x-qr$

So $p=-p \Rightarrow p = 0$ and $q = -q \Rightarrow q = 0$
Title: Re: sketching graph + 2 functions questions
Post by: TrueTears on December 04, 2008, 04:10:19 pm: So, how would u define the domains here?

$f(x)=\begin{cases} x^2-4 for x\in(-\infty,2) \\ x for x\in[2,\infty)\end{cases}$

if $h(x)= 2x$ find $h(f(x))$ .

The $dom h$ is just R right? so the domains of f(x) won't change when u plug it in the hybrid equation of $h(f(x))$ right?
Title: Re: sketching graph + 2 functions questions
Post by: /0 on December 04, 2008, 04:20:48 pm: Yeah the maximal domain for both functions (and consequently the composite function) is R.