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Let $f(x)$ be a differentiable function such that $f'(-x) = -f'(x)$ show that $f(-x) = f(x)$.

Unfortunately I am not sure what to do with this simple looking problem. I was having trouble trying to draw any useful conclusions about the function given the property of the derivative... I guess it would've been easier to go the other way around. How is this problem solved?

Thank you in advance for any help!

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What about integrating? –  Giuseppe Negro Apr 16 '11 at 12:01
    
This should feel intuitive to you. Can you see how this works when $f$ is a polynomial? A power series? –  Qiaochu Yuan Apr 16 '11 at 17:27
    
@Qiaochu: thanks for that comment, I guess the derivative of a polynomial satisfies $f'(-x)=-f'(x) \Rightarrow f'$ is odd with degree $n \Rightarrow \int f'(x)dx$ is even with degree $n+1$..? Unfortunately I can't say much about a power series or a differentiable function in general... –  ghshtalt Apr 16 '11 at 17:36
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3 Answers 3

up vote 3 down vote accepted

Let

g(x) = f(x) - f(-x)

h(x) = f(x) + f(-x)

Then

f(x) = (g(x) + h(x))/ 2

g'(x) = f'(x) + f'(-x) = 0

so g(x) = C, a constant

h'(x) = f'(x) - f'(-x) = 2f'(x)

So

f'(x) = (g'(x) + h'(x)) / 2 = C / 2 + f'(x)

=> C = 0, and so g(x) = f(x) - f(-x) = 0 => f(x) = f(-x)

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Define $g(x) = f(x)-f(-x)$. Your goal is then to prove that $g(x)=0$, and the information you have on $f'$ probably says something useful about $g'$.

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thank you for this answer. Unfortunately I can't manage to finish the problem with it yet. $g(x) = f(x)-f(-x) \Rightarrow g'(x)=f'(x)+f'(-x) \Rightarrow g'(x) =0 \Rightarrow g(x)=C \Rightarrow ?$ is the rest pretty much as Chris shows, or did you have something else in mind? (The only thing is that I don't know how I would have defined all these other functions and so on...) –  ghshtalt Apr 16 '11 at 13:04
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@ghshtalt : what about $g(0)$ ? –  mercio Apr 16 '11 at 13:12
    
$g(0) = f(0)-f(-0)=0$ ? –  ghshtalt Apr 16 '11 at 13:26
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I presume this is homework, so I'll just give a hint: What happens if you integrate both sides of the equation $f'(-x)=-f'(x)$?

To really see what happens, you can look at polynomials as well. What are the odd polynomials, and what are their primitives?

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@Thomas: It's not homework, but thanks for the hint anyway. If I integrate both sides of the above equation I get $f(-x) + C = -\int f(x)dx$ right? I am not sure what to do with that and in I general get thrown off by the constant whenever I try to use integration... –  ghshtalt Apr 16 '11 at 12:12
    
@ghshtalt: there was a small typo in the post which I corrected. You should integrate $\int_0^x f'(-x') dx' = -\int_0^x f'(x') dx'$. What can you deduce? –  Fabian Apr 16 '11 at 12:18
    
@Fabian: Thank you, I noticed that as well, but wasn't sure if it would turn out to be a 'trick.' As for deducing something: what more than $f(-x) = -f(x)$ or maybe even $f(-x)+f(x)=0$ should I see? Also, why is it ok to evaluate the definite integral here? (sorry if i'm missing very obvious stuff!) –  ghshtalt Apr 16 '11 at 12:28
    
@ghshtalt: I guess you did not integrate correctly. The equation after performing the integral should read $f(0) -f(-x)= f(0) -f(x)$. –  Fabian Apr 16 '11 at 12:34
    
@user6312: why $f(-x)\neq f(x)$? –  Fabian Apr 16 '11 at 12:36
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