Suppose $f : (0, 1) → \mathbb R$ is differentiable on $(0, 1)$ and that for $x,y ∈ (0,1)$ with $x < y$ we have that $f′(x) < 0$ and $f′(y) > 0$. Show that there exists $z ∈ (x, y)$ such that
$f(z) = \inf${$f(a) : a ∈ [x,y]$} and $f′(z) = 0.$
I can picture this problem really easily but just have no idea on how to get started on paper. Maybe something to do with IVT? Or the MVT instead? Would love some help with it. Sorry if there is formatting/tag errors it's my first post on here :)
Since $f'(x) <0$ implies that $f(x+h) -f(x) >0$ for $|h| <\delta$ for some $\delta >0$, we see that the infimum is not attained in $x$. Similarly, because of $f'(y) >0$, we can deduce that the infimum is not attained in $y$.
Now, let $z \in (x,y)$ be a global minimum of $f$ on $[x,y]$. That implies $f(z) \leq f(z+h)$ for all $|h| < \delta$ for a fixed $\delta>0$. Taking $h >0$ we get $$0 \leq \frac{f(z+h)-f(z)}{h}$$ and thus $0 \leq f'(z)$. On the other hand, we have $$0 \geq \frac{f(z+h)-f(z)}{h}$$ for negative $h>0$ with $|h| < \delta$. This shows $ 0 \geq f'(z)$.