Proofs of Derivative Applications Facts — Question 2

Question 2 - Solution

Since $f$ is continuous on the closed interval $[a,b]$, the Extreme Value Theorem guarantees that $f$ attains both a maximum and a minimum on $[a,b]$.

Let $$M=\max_{x\in[a,b]} f(x), \qquad m=\min_{x\in[a,b]} f(x).$$

If $M=m$, then $f$ is constant on $[a,b]$, and hence $$f'(x)=0 \quad\text{for all }x\in(a,b).$$ In particular, there exists $c\in(a,b)$ such that $f'(c)=0$.

Now assume that $M\neq m$. Since $f(a)=f(b)$, the maximum or minimum cannot occur at both endpoints.

Thus, at least one of the extreme values $M$ or $m$ must occur at some point $c\in(a,b)$.

At such a point $c$, the function $f$ has either a local maximum or a local minimum. Because $f$ is differentiable at $c$, the derivative must vanish there.

Therefore, $$f'(c)=0.$$

$$\boxed{\text{There exists }c\in(a,b)\text{ such that }f'(c)=0}$$