$f$ is differentiable convex function and $x_0$ is a critical point for f

Asked Dec 16 '20 at 19:20

Active Dec 16 '20 at 19:20

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I have the following theorem:
Let $f:(a,b)\rightarrow \mathbb{R}$ be differentiable function. Then $f$ is convex if and only if
$f(y)\geq f(x)+f'(x)(y-x)$ for every $x,y\in (a,b)$

Then i am given the following:
Suppose that $f:(a,b)\rightarrow \mathbb{R}$ is a differentiable convex function and $x_0\in (a,b)$ is a critical point for $f$. What can you say about $x_0$ using the theorem.

This seems like a pretty simple question but i'm honestly not sure what i'm supposed to answer. Anyone got any ideas what might be good to say here? anything is appreciated

asked Dec 16 '20 at 19:20

Chongyun

Do you have any ideas? What does it mean for $x_{0}$ to be a critical point? – DMcMor Dec 16 '20 at 19:23
1

If $f'(x_0) = 0$ then $f(y) \ge f(x_0)$ for all appropriate $y$. – copper.hat Dec 16 '20 at 19:23

$f$ is differentiable convex function and $x_0$ is a critical point for f

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