# Applications of the Fundamental Theorem of Calculus

$$g(20)=0$$
$$g'(t)\geq 0$$ 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?

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

$$g'(t)\ge 0$$ 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 $$t= 20$$.