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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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    @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...2011-04-16

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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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    @Fabian: Too early in the morning (west coast).2011-04-16
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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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    $g(0) = f(0)-f(-0)=0$ ?2011-04-16
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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)