$g'(t)\geq 0$ for all values of $t$.
The function is differentiable and satisfies the conditions above. Let $F$ be the function given by $F(x)=\int_0^x g(t) dt$. What must be true?
$F$ has a local minimum at $F=20$.

This is the question and answer that I was given. (It was a multiple choice question, I just included the correct answer.) I know this is correct, but why is it so? How do I understand this based on the given information?

  • 1
    $\begingroup$ $F'(x) = g(x)$ by the fundamental theorem, so that $F'(20) = g(20) = 0$. Can you go from here? $\endgroup$
    – Freddie
    Aug 1, 2019 at 3:47

2 Answers 2


$g'(t)\ge 0$ for all $t$ implies $g(t)$ is monotone increasing. $g(20)=0$ implies $g(t)\le 0$ for $t\lt 20$. This implies $F(x)$ is monotone decreasing for $x\lt 20$. $F(x)$ stops decreasing at $20$ and may start to increase, therefore $F(x)$ has a minimum at $20$.


The claim follows directly using the Taylor polynomial of degree $2$ which can be applied since $g$ is assumed to be differentiable.

So, for any $h \neq 0$ there is a $\tau_h$ between $20$ and $20+h$ such that

$$F(20+h) = F(20) + \underbrace{F'(20)}_{=g(20) =0}h + \underbrace{F''(\tau_h)}_{=g'(\tau_h)}h^2 = F(20) + \underbrace{g'(\tau_h)}_{\geq 0}h^2 \geq F(20)$$

Hence, $F$ has a local minimum at $t= 20$.


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