Prove that the function has at least one stationary point in the set $(-1,1)$

Question

Given a function $f:[-1,1]{\rightarrow\mathbb{R}}$ such that $f$ is continuous on $[-1,1]$ and differentiable on $(-1,1)$ and such that$$f(-1)=0, \space f(0)=-1, \space f(1)=2.$$ Prove that the function has at least one stationary point in the set $(-1,1)$

The function is continuous in a closed bounded interval, then we can apply Weierstrass Theorem which implies that the function has the maximum and the minimum in the domain $[−1, 1].$

Not too sure where to go from here.

Answer 1

You can actually prove more about this stationary point: it must be a local (and absolute) minimum. By compactness, the function has an absolute minimum. By the endpoint knowledge and that $f(0)$ is smaller than both, this min cannot be at the ends. Hence the min must be a local min, so $f'(c)=0$ at this point.

Answer 2

Another way to think about it is that you can use the mean value theorem on $[-1,0]$ to see that the derivative is negative somewhere in $(-1,0)$. You can then apply the mean value theorem again on $[0,1]$ to see that the derivative is positive somewhere in $(0,1)$. Since derivatives satisfy the intermediate value property (this is called Darboux's theorem) it follows that the derivative must have a zero somewhere in $(-1,1)$.

Answer 3

Because $f$ continuous it has Darboux property, therefore $f(0)=-1, \space f(1)=2$ implies there is $x_0 \in (0,1)$ such that $f(x_0) = 0$. Now apply Rolle theorem to $f$ restricted to $[-1, x_0]$