The problem says:

What is $f'(0)$, given that

$f\left(\sin x −\frac{\sqrt 3}{2}\right) = f(3x − \pi) + 3x − \pi$, $x \in [−\pi/2, \pi/2]$.

So I called $g(x) = \sin x −\dfrac{\sqrt 3}{2}$ and $h(x)=3x − \pi$.

Since $f(g(x)-h(x))=3x − \pi$, I called $g(x)-h(x) = j(x) = \sin x −\frac{\sqrt 3}{2} - 3x +\pi$

I imagined that I should find the function $f$, by finding the inverse of $j(x)$ and doing $(f \circ j \circ j^{-1})(x)$ but I discovered that it is too much difficult.

I guess that's not the best way.

What should I do?

  • $\begingroup$ But $f(g(x) - h(x))$ is not $3x - \pi$. You can only say that $f(g(x)) - f(h(x)) = 3x - \pi$. $f(a) - f(b) \ne f(a - b)$ (in general). $\endgroup$ – M. Vinay Apr 26 '16 at 5:22
  • $\begingroup$ OMG, that's true. So I really don't know where to go $\endgroup$ – Carlos Mendes Apr 26 '16 at 5:32
  • $\begingroup$ Just differentiate, use chain rule, and then try to get $f'(0)$ using an appropriate value of $x$. $\endgroup$ – M. Vinay Apr 26 '16 at 5:47

Given: $$f(\sin x - \frac{\sqrt 3}{2}) = f(3x-\pi) + 3x - \pi$$ Taking derivatives on both sides w.r.t. $x$, we get, $$f'(\sin x - \frac{\sqrt 3}{2}).\cos x = 3.f'(3x-\pi) + 3$$ Put $x=\frac{\pi}{3}$, we get, $$f'(\sin \frac{\pi}{3} - \frac{\sqrt 3}{2}).\cos \frac{\pi}{3} = 3.f'(3.\frac{\pi}{3}-\pi) + 3$$ $$f'(0).\frac{1}{2} = 3.f'(0) + 3$$ Solving for $f'(0)$, we get, $$f'(0)=-\frac{6}{5}$$


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