1

Lets's say we have a differentiable function

$f:[a,b]\to \mathbb{R}$

with

$f^\prime\equiv0$

How do I show that

$f\equiv C$

by using the mean value theorem?

5xum
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ViktorG
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  • Have you searched before asking? Key terms: constant, derivative zero, mean value. –  Jan 07 '15 at 17:02
  • Yes I did but it seems that i may have had a different way of thinking and didn't notice that a similar question has been asked. – ViktorG Jan 07 '15 at 17:05

2 Answers2

4

One approach is to prove by contradiction (or contrapositive).

Suppose that $f$ is not constant. Then, there exists a $b' \in [a,b]$ such that $f(b') \neq f(a)$. Apply the mean value theorem on the interval $[a,b']$. What do you conclude?

Ben Grossmann
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  • I seem to struggle with the mean value theorem. $\exists b'\in [a,b] : f(b') \neq f(a)$ If I apply the mean value theorem now i have : $\frac{f(b')-f(a)} {b'-a}$ How do I continue? – ViktorG Jan 07 '15 at 12:51
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    What does the mean value theorem say, in words? The statement "I have $\frac{f(b')-f(a)}{b'-a}$" doesn't mean anything to me as written. But what does the mean value theorem tell you about that number? – Ben Grossmann Jan 07 '15 at 12:56
  • It says that there is a point which has the average slope of the whole function in the $[a,b]$ , doesn't it? – ViktorG Jan 07 '15 at 13:01
  • Something like that. The mean value theorem states *there exists a point $c \in [a,b']$ such that $f'(c) = \frac{f(b') - f(a)}{b' - a} \neq 0$.* – Ben Grossmann Jan 07 '15 at 13:22
  • I think i got it. Thank you :) – ViktorG Jan 07 '15 at 13:24
4

For $x\in (a,b]$, we have $$\tag{1} f(x)-f(a)=f'(\zeta)\cdot(x-a)\mbox{ for some }\zeta\in (a,x)$$ by the mean value theorem. Now, since $f'\equiv 0$, it follows from $(1)$ that $$f(x)-f(a)=0,$$ that is $f(x)=f(a)$.

Paul
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