1. ## Explicit Expression?

Hello, the full question is: Consider the function f(x)=|x2-4| A) Find the 2 points at which the function is not differentiable [SOLVED] B) For the larger of the two points in (a), find the left and right limit of the slope of the function [SOLVED] ...The part I'm having trouble with:

C) Give an explicit expression of f ' whever it is defined.

Anyone have any idea how to do this part?

2. ## Re: Explicit Expression?

Consider how you can define f(x) analytically in the different points (x < -2, -2 <= x < 2, x >= 2). You will have three different functions analytic across those regions and not differentiable at certain points, but differentiable everywhere else.

3. ## Re: Explicit Expression?

Hello, the full question is: Consider the function f(x)=|x2-4| A) Find the 2 points at which the function is not differentiable [SOLVED] B) For the larger of the two points in (a), find the left and right limit of the slope of the function [SOLVED] ...The part I'm having trouble with:

C) Give an explicit expression of f ' whever it is defined.

Anyone have any idea how to do this part?
The absolute value can be written as

$\displaystyle |x|=\sqrt{x^2}$

So your function can be written

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

This is the composition of three functions

$\displaystyle u=x^2-4 \quad v=\sqrt{u} \quad w=v^2$

So by the chain rule

$\displaystyle f'(x) = \frac{dw}{dv}\frac{dv}{du}\frac{du}{dx}$

Can you finish from here?

4. ## Re: Explicit Expression?

I think that a piecewise definition qualifies as "an explicit expression of f' wherever it is defined". There will be three regions of definition, $\displaystyle x<-2$, $\displaystyle -2<x<2$, and $\displaystyle x>2$. The derivative is undefined at 2 and -2.

- Hollywood

5. ## Re: Explicit Expression?

Consider the function f(x)=|x2-4|
[B]C) Give an explicit expression of f ' whever it is defined.
I would write it this way:
$\displaystyle f'(x) = \left\{ {\begin{array}{rl} {2x,} & {\left| x \right| > 2} \\ { - 2x,} & {\left| x \right| < 2} \\\end{array}} \right.$

6. ## Re: Explicit Expression?

Good point - two of my regions have the same expression for f(x) (and therefore also for f'(x)).

- Hollywood

7. ## Re: Explicit Expression?

Originally Posted by TheEmptySet
The absolute value can be written as

$\displaystyle |x|=\sqrt{x^2}$
I hate to reply to myself but I just relized that I wrote the composition in the wrong order, So I will correct it in the post!

It should have been

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

This is the composition of three functions

$\displaystyle u=x^2-4 \quad v=u^2 \quad w =\sqrt{v}$

Now if we take the derivative we get

$\displaystyle f'(x)=\frac{dw}{dv}\frac{dv}{du}\frac{du}{dx} = \frac{1}{2\sqrt{v}}(2u)(2x)$

Subbing back out we get

$\displaystyle f'(x)=\frac{u}{\sqrt{u^2}}2x=\frac{2x(x^2-4)}{\sqrt{(x^2-4)^2}}=\frac{2x(x^2-4)}{|x^2-4|}$

Note this will match up with the peicewise defintion and you can also see where it is undefined as well.

Again sorry about the error in my first post.

TES

8. ## Re: Explicit Expression?

Originally Posted by TheEmptySet
I hate to reply to myself but I just relized that I wrote the composition in the wrong order, So I will correct it in the post!

It should have been

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

This is the composition of three functions

$\displaystyle u=x^2-4 \quad v=u^2 \quad w =\sqrt{v}$

Now if we take the derivative we get

$\displaystyle f'(x)=\frac{dw}{dv}\frac{dv}{du}\frac{du}{dx} = \frac{1}{2\sqrt{v}}(2u)(2x)$

Subbing back out we get

$\displaystyle f'(x)=\frac{u}{\sqrt{u^2}}2x=\frac{2x(x^2-4)}{\sqrt{(x^2-4)^2}}=\frac{2x(x^2-4)}{|x^2-4|}$

Note this will match up with the peicewise defintion and you can also see where it is undefined as well.

Again sorry about the error in my first post.

TES
Thank you
My question is, it asks for this expression wherever x is defined, so should I make a statement after giving this equation saying |x| > 2 and -|x| < 2 (because those values would produce asymptotes) ?

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### explicit expression m

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