The given function is
$ f(x) = |9-x^2| $ on $[-3,3]$;
also find point where the derivative is zero.
I got the value of $c$ as $\pm3$, but according to the theorem the value should be in between $-3$ and $3$, i.e., excluding the $-3$ and $3$.
[Math] Verify Rolle’s theorem.
calculusderivativesmaxima-minimameans
Best Answer
The function $|9-x^2|$ is not differentiable on $-3$ and $3$, so the derivatives cannot be zero at that point, as they do not exist at all. You must have made a mistake somewhere.
In your problem particularly, as $f(x)=9-x^2$ we have $f'(x)=-2x$, as $9-x^2 \ge 0$ on $[-3,3]$, the point where the derivative is $0$ is $0$, and only $0$.
Also, note that Rolle's Theorem says that the point where the derivatives are zero are all in $[a,b]$ where $f(a)=f(b)$. It merely says that there exists such a point.
For example, take $f(x)=0$ on $[1,3]$. There exists a value where $f'(x)=0$ on $(1,3)$, but $f'(1)=f'(3)=0$ as well. So values where $f'(x)=0$ need not always exist in the interval $(a,b)$.