# Thread: verification of rolle's theorem

1. ## verification of rolle's theorem

Verify rolle's theorem for |x| in [-1 1]

2. Originally Posted by ramya
Verify rolle's theorem for |x| in [-1 1]
Rolle's Theorem states as follows:

Suppose $f(x)$ is a function that satisfies all of the following:

1. $f(x)$ is continuous on the closed interval $\,[a,\, b]$
2. $f(x)$ is differentiable on the open interval $\,(a,\, b)$
3. $f(a)\, =\, f(b)$

Then there is a number $\,c\,$ such that $\,a\, <\, c\, <\, b\,$ and $\,f'(c)\, =\, 0.$ Or, in other words $\,f(x)\,$ has a critical point in $\,(a,\, b)$.
I will assume that the exercise actually told you to work with something like "$f(x)\, =\, |\,x\,|$" on the given interval.

That said, how far have you gotten in applying the formulaic statement of the Theorem? In particular, what did you do with part (2) of the above statement?

3. ## Not differentiable

The graph of mod(x) has a sharp turn at x=0 which means we can not draw a unique tangent at that point. This makes it non- differentiable at that point.

Aliter: the differential of mod(x) = mod (x) / x , wherever it is defined. This can be proved from first principles. This differential is undefined at x=0.

Hence, Rolle's theorem can not be applied.

4. Originally Posted by Giridhar
The graph of mod(x) has a sharp turn at x=0 which means we can not draw a unique tangent at that point. This makes it non- differentiable at that point.

Aliter : the differential of mod(x) = mod (x) / x , wherever it is defined. This can be proved from first principles. This differential is undefined at x=0.

Hence, Rolle's theorem can not be applied.
.

I don't know what those words and that one phrase (that I highlighted in bold) mean in the context of your post.

If f(x) = mod(x) is supposed to be the absolute value function, $\ \ \ \ \ \ \$ Edit: Please see the post directly below mine.

If f(x) = |x|, $\ \$ then f'(x) = x/|x|

Here are some supporting steps:

f(x) $= |x| \ \ \ \implies$

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

$f(x) = (x^2)^{\frac{1}{2}}$

f'(x) $\ = \ \frac{1}{2}(x^2)^{ \frac{-1}{2}}(2x) \ \ \ \implies \ \ \ \ \ \ \ \ \ \$ (There is a faint "-1" in the fractional exponent -1/2.)

$f'(x) \ = \ \dfrac{x}{(x^2)^{\frac{1}{2}}} \ \ \ \implies$

$f'(x) \ = \ \dfrac{x}{|x|} \ \ \ \ \ \ \ \ \$Edit: Please see the post directly below mine.

5. Originally Posted by lookagain
If f(x) = mod(x) is supposed to be the absolute value function, then f'(x) does not equal (mod(x))/x. It's the "flip" of it.

If f(x) = |x|, $\ \$ then f'(x) = x/|x|
But $\dfrac{x}{|x|} \equiv \dfrac{|x|}{x}$

for any $x \ne 0$, and is undefined if $x = 0$

6. Another way of saying this is that the derivitive of |x| is 1 if x> 0, -1 if x< 0, undefined at x= 0.

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