I think I was/am missing a key concept of Rolle's theorem.
The question says find the value(s) of c that satisfies Rolle's theorem for
$y=-x^2+2$; $[-2,2]$
I see that this is simply a parabola with vertex at $(0,2)$ and that at that point there is possible a horizontal line (slope$=0$).
The theorem states that if $f$ is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$ and $f(a)=0 $ and $f(b)=0$, then there is at least one point c in the interval $(a,b)$ such that $f'(c)=0$.
So, I first tested to see if $f(-2)=0$ and it does NOT. And further $f(2)$ does NOT equal $0$, so I assumed that I could go no further.
Is that correct? I think maybe what I should do is find the roots of $f(x)$ and if those roots are within $[-2,2]$ [and they ARE], THEN find c.
A little confused on this key concept of Rolle's. Maybe it was a misprint and they meant to put the interval as $[-\sqrt (2), \sqrt (2)]$ ?
Do the endpoints have to be used in $f(a)$????
Best Answer
There is a generalization of Rolle theorem which says that if $f(a)=f(b)$ there exists $c$ such that $f'(c)=0$. Here $f(-2)=f(2)$. This can be easily seen from the previous case, suppose that $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ and $f(a)=f(b)$. Define $g$ on $[a,b]$ by $g(x)=f(x)-f(a)$, $g(a)=g(b)=0$ and there exists $c$ such that $g'(c)=f'(c)=0$.
https://en.wikipedia.org/wiki/Rolle%27s_theorem