Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
up vote 3 down vote accepted


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

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


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)


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)

share|cite|improve this answer

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 $g'$.

share|cite|improve this answer
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
@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

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?

share|cite|improve this answer
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.