2

This is what i have:

Put $g(t) = f(t) - ct$.

Then $g'(a)<0$ so that $g(t_{1}) < g(a)$ for some $t_{1} \in (a,b)$ and $g'(b)>0$ so that $g(t_{2}) < g(b)$ for some $t_{2} \in (a,b)$.

Hence $g$ attains its minimum on $[a,b]$ at some point x such that $a<x<b$.

Then, $g'(x)=0$, hence $f'(x)=c$

kahen
  • 15,760
user77107
  • 819
  • Please read the description of the tags before you add them to your question. This has absolutely nothing to do with the maximum (modulus) principle. – kahen May 16 '13 at 00:48
  • Some might consider it terse (for presumably an introductory analysis course); but it looks fine to me. Perhaps at the outset point out that $g$ does have a minimum on $[a,b]$ and state why. Also, you might explain why the observations made in the second paragraph hold and explicitly state that they imply the minimum in fact occurs at a point in $(a,b)$. – David Mitra May 16 '13 at 01:06

2 Answers2

1

This is known as Darboux's theorem. You need to do a bit more work because it could happen that you have to pick $x=a$ or $x=b$.

kahen
  • 15,760
  • It looks to me that the OP took care of the endpoints. He states, e.g., "$g'(a)<0$ so that $g(t_1)<g(a)$ for some $t_1\in(a,b)$" (so the minimum is not achieved at $a$). – David Mitra May 16 '13 at 00:59
0

This is known as the Intermediate Value Theorem. You can find a proof worked out searching Google or Youtube.

Also check out Darboux's theorem, which states:

Let ${I}$ be an open interval, $f\colon I\to \mathbf{R}$ a real-valued differentiable function. Then $f'$ has the intermediate value property: If $a$ and $b$ are points in $I$ with $a\leq b$, then for every $y$ between $f'(a)$ and $f'(b)$, there exists an x in [a,b] such that $f'(x)=y$

Nate
  • 1