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$f:\mathbb{R} \rightarrow \mathbb{R}$ is such that $f'(x)$ exists $\forall x.$

And $f'(-x)=-f'(x)$

I would like to show $f(-x)=f(x)$

In other words a function with odd derivative is even.

If I could apply the fundamental theorem of calculus

$\int_{-x}^{x}f'(t)dt = f(x)-f(-x)$ but since the integrand is odd we have $f(x)-f(-x)=0 \Rightarrow f(x)=f(-x)$

but unfortunately I don't know that f' is integrable.

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Let $g(x)=f(-x)$. Then $g'(x)=-f'(-x)=f'(x)$.

Since $g(0)=f(0)$ and $g'=f'$, it follows from the mean value theorem that $g=f$.

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f and g are both equal to $g'(c)x+g(0)$ – user9352 Jan 11 '12 at 0:54
@user9352: What is $c$? Note that $f$ is could be an arbitrary even function, so your formula is incorrect if $c$ is a constant. If say $f(x)=x^2$, then your formula says $x^2=2cx$, so in that case $c=\frac{1}{2}x$? – Jonas Meyer Jan 11 '12 at 0:55
yeah i guess c depends on x i just got that from the mvt, trying to see how the result follows. c is the mysterious number between 0 and x, $f(x)-f(0)=f'(c)x$ – user9352 Jan 11 '12 at 2:23
@user9352: Antiderivatives of functions on $\mathbb R$ are unique (if they exist) up to an added constant. The easiest way to apply the MVT is to the function $h(x)=g(x)-f(x)$. I recommend contraposition (or contradiction): If $h(0)=0$ and $h(a)\neq 0$ for some $a$, then $h'(c)\neq 0$ for some $c$. – Jonas Meyer Jan 11 '12 at 2:28
  • Define functions $f_0(x)=(f(x)+f(-x))/2$ and $f_1(x)=(f(x)-f(-x))/2$. Then $f_0$ and $f_1$ are also differentiable, and $f_0$ is even and $f_1$ is odd.

  • Show that the derivative of an odd function is even, and that of an even function is odd.

  • From the equality $f'=f_0'+f_1'$ conclude that $f_1$ is constant and, therefore, zero.

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Don't you mean $f_0$ is even and $f_1$ is odd? – M Turgeon Jan 11 '12 at 1:40
Yeah, that. I even picked the indices in $\mathbb Z/2$ and all! – Mariano Suárez-Alvarez Jan 11 '12 at 2:36

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