Author Topic: TyErd's questions (Read 42765 times) Tweet Share

TyErd · « **Reply #150 on:** May 03, 2010, 04:50:24 pm »

For the graph of $y= 4x^3 + 27x^2 - 30x +10$ the subset of R for which the gradient is negative is given by the interval........?

TyErd · « **Reply #151 on:** May 03, 2010, 04:56:03 pm »

Differentiate $f(x) = |2x+4|$ . What are the steps to differentiating a modulus function.

the.watchman · « **Reply #152 on:** May 03, 2010, 05:05:49 pm »

Quote from: TyErd on May 03, 2010, 04:56:03 pm

Differentiate $f(x) = |2x+4|$ . What are the steps to differentiating a modulus function.

You can use a chain rule approach, or this:

When the inside of the mod is positive (eg. when $2x+4\geq 0$ ), then the derivative is [2x-4]'

When the inside of the mod is positive (eg. when $2x+4<0$ ), then f(x) becomes -2x-4, so the derivative is [-2x-4]'

Hence the derivative can be written as a hybrid function according to the above

the.watchman · « **Reply #153 on:** May 03, 2010, 05:08:58 pm »

Quote from: TyErd on May 03, 2010, 04:50:24 pm

For the graph of $y= 4x^3 + 27x^2 - 30x +10$ the subset of R for which the gradient is negative is given by the interval........?

Well, first off, $\frac{dy}{dx} = 12x^2 + 54x - 30$

For the question, $\frac{dy}{dx} < 0$

So $12x^2 + 54x - 30<0$

$2x^2 + 9x - 5<0$

$(2x-1)(x+5)<0$

Solve either graphically or algebraically to get:

$-5<x<\frac{1}{2}$

TyErd · « **Reply #154 on:** May 03, 2010, 05:15:05 pm »

Quote from: the.watchman on May 03, 2010, 05:08:58 pm

Quote from: TyErd on May 03, 2010, 04:50:24 pm
For the graph of $y= 4x^3 + 27x^2 - 30x +10$ the subset of R for which the gradient is negative is given by the interval........?

Well, first off, $\frac{dy}{dx} = 12x^2 + 54x - 30$

For the question, $\frac{dy}{dx} < 0$

So $12x^2 + 54x - 30<0$

$2x^2 + 9x - 5<0$

$(2x-1)(x+5)<0$

Solve either graphically or algebraically to get:

$-5<x<\frac{1}{2}$

okay I get that part except how come when i solve algebraicaly i get x<-5. When i do it graphically i get it right.
and the modulus one im still very confused

m@tty · « **Reply #155 on:** May 03, 2010, 05:35:09 pm »

$f(x)=|2x+4|=\sqrt{(2x+4)^2}$

Or you can say

$f(x)=\begin{cases} 2x+4, x\geq -2 \\ -2x-4, x<-2\end{cases}$

The piece wise defined function is much easier to differentiate. As you don't have to use the chain rule twice...

$f'(x)=\begin{cases}\frac{d}{dx}(2x+4), x> -2 \\ \frac{d}{dx}(-2x-4), x<-2\end{cases} = \begin{cases}2, x>- 2 \\ -2, x<-2\end{cases}$

NOTE: There is no value for the derivative at $x=-2$ .

EDIT: The other method goes like this... (I stupidly used this method in the exam last year and lost a mark..

)

$f(x)=\sqrt{(2x+4)^2}$

$f(x)=\sqrt{u}$ where $u=(2x+4)^2$

You could 'officially' use the chain rule again as in write it down on paper, or you could just do it quickly in your head.

$f'(x)=\frac{1}{2\sqrt{u}}\cdot u'$

and

$u'=4(2x+4)^1=8x+16$

So $f'(x)=\frac{8x+16}{2\sqrt{(2x+4)^2}}=4\cdot \frac{x+2}{2|x+2|}=2\cdot \frac{x+2}{|x+2|}$ .

And $\frac{x+2}{|x+2|}=\begin{cases}1, x> -2 \\ -1, x<-2\end{cases}$
(This is because they are always both the same magnitude, and the denominator is always positive, so the sign is determined by the sign of the numerator, which is positive where $x>-2$ and negative where $x<-2$ )

Hence $2\cdot \frac{x+2}{|x+2|}=\begin{cases}2\cdot 1 , x>-2 \\ 2\cdot -1 , x<-2\end{cases} = \begin{cases}2,x>-2 \\ -2, x<-2\end{cases}$ .

TyErd · « **Reply #156 on:** May 03, 2010, 05:37:19 pm »

thankyou! i get it now

the.watchman · « **Reply #157 on:** May 03, 2010, 05:49:37 pm »

Algebraic is more complicated than you think, there are two cases to consider

CASE 1:
(2x-1)>0 AND (x+5)<0 (pos. times neg. = neg.)

CASE 2:
(2x-1)<0 AND (x+5)>0 (neg. times pos. = neg.)

Try it from there!

Quote from: m@tty on May 03, 2010, 05:35:09 pm

$f(x)=|2x+4|=\sqrt{(2x+4)^2}$

Or you can say

$f(x)=\begin{cases} 2x+4, x\geq -2 \\ -2x-4, x<-2\end{cases}$

The piece wise defined function is much easier to differentiate. As you don't have to use the chain rule twice...

$f'(x)=\begin{cases}\frac{d}{dx}(2x+4), x\geq -2 \\ \frac{d}{dx}(-2x-4), x<-2\end{cases} = \begin{cases}2, x\geq 2 \\ -2, x<-2\end{cases}$

BUT the function is not differentiable at x=2 (for the derivative, pos and neg limits different)

m@tty · « **Reply #158 on:** May 03, 2010, 05:50:31 pm »

Quote from: the.watchman on May 03, 2010, 05:49:37 pm

BUT the function is not differentiable at x=-2 (for the derivative, pos and neg limits different)

YEAH I fixed that up. Thanks.

TyErd · « **Reply #159 on:** May 03, 2010, 06:16:22 pm »

why is there no value for the derivative at x=-2

ZachCharge · « **Reply #160 on:** May 03, 2010, 06:25:17 pm »

Quote from: TyErd on May 03, 2010, 06:16:22 pm

why is there no value for the derivative at x=-2

At x=-2 the function is at a cusp, which are impossible to get the gradient function of.

TyErd · « **Reply #161 on:** May 03, 2010, 06:37:40 pm »

Quote from: m@tty on May 03, 2010, 05:35:09 pm

$f(x)=|2x+4|=\sqrt{(2x+4)^2}$

Or you can say

$f(x)=\begin{cases} 2x+4, x\geq -2 \\ -2x-4, x<-2\end{cases}$

The piece wise defined function is much easier to differentiate. As you don't have to use the chain rule twice...

$f'(x)=\begin{cases}\frac{d}{dx}(2x+4), x> -2 \\ \frac{d}{dx}(-2x-4), x<-2\end{cases} = \begin{cases}2, x>- 2 \\ -2, x<-2\end{cases}$

NOTE: There is no value for the derivative at $x=-2$ .

What happened to to the greater than or equal too symbol. Why did you make it just greater than -2?

Potter · « **Reply #162 on:** May 03, 2010, 07:16:44 pm »

Hey guys,

Sorry TyErd for crashing your thread, but I didn't really see the point of making a new thread.

Had a sac last week and my friend and I are in a disagreement with what is right.. Just thought you guys might be able to help.
So, here we go.

y=e^(x+2) is translated 3 units to the right

So, y=e^(x-1)

The next question asked us if the graph is reflected in the y axis, what is the new equation?

Is it y=e^(-x-1)

OR

y=e^-(x-1)

Thanks guys.

TyErd · « **Reply #163 on:** May 03, 2010, 07:20:09 pm »

m@tty · « **Reply #164 on:** May 03, 2010, 07:21:11 pm »

As said above, there is no value for the derivative at x=-2, as there is a cusp. At a cusp you don't have one value for the gradient at this point, you can draw more than one tangent. Or, more correctly, the left and right hand limits are not equal.

let $f(x)=|2x+4|$

To find the gradient using first principles you take the limit:

$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

And for this limit to exist, the left and right hand limits must be equal.

Firstly, the left hand limit:

$\lim_{h\to 0^-}\frac{f(-2+h)-f(-2)}{h}=\lim_{h\to 0^-}\frac{|2(-2+h)+4|-|2(-2)+4|}{h}$

$=\lim_{h \to 0^-} \frac{|2h|-|0|}{h}$

$=\lim_{h \to 0^-} \frac{-2h}{h}$ (as $h<0$ )

$=-2$

And the right hand limit:

$\lim_{h \to 0^+} \frac{f(-2+h)-f(-2)}{h}=\lim_{h\to 0^+} \frac{|2h|}{h}$

$=\lim_{h\to 0^+} \frac{2h}{h}=2$

As these limits are not equal, the derivative is not defined at this point, $x=-2$ .

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