let $f$ is a function and satisfy in the equation $\cos(f(x))+ \frac{8}{\pi^3}(f(x))^3 +2f(x)=1+x$

Question

let $f$ is a function and satisfy in the following equation :$$\cos(f(x))+ \frac{8}{\pi^3}(f(x))^3 +2f(x)=1+x$$

find $2f^{\prime}(0)+f^{\prime}(\pi)$ .

I find $f(0)=0$ and with derivation ( if we let $f$ is continuous in $0$ and $\pi$) we have :$$(\cos(f(x))+ \frac{8}{\pi^3}(f(x))^3 +2f(x))^{\prime} =-f^{\prime}(x)\sin(f(x))+3f^{\prime}(x) \frac{8}{\pi^3}(f(x))^2 +2f^{\prime}(x)=1$$ and so we can compute $f^{\prime}(0)=\frac{1}{2}$

Answer 1

HINT : I suppose that it is an academic exercise. If so, the equation to be solve is not too complicated. In the first equation, all difficulty to solve it for $f(x)$ comes from the $\cos(f(x))$. One can guess that this $\cos$ term is nul for $x=\pi$ , making the equation easy to solve.

So, we try $f(\pi)=\frac{\pi}{2}$ : $$\cos(\frac{\pi}{2})+ \frac{8}{\pi^3}(\frac{\pi}{2})^3 +2\frac{\pi}{2}=1+\pi$$ By luck, it agrees. Thus : $$f(\pi)=\frac{\pi}{2}$$ I suppose that you can continue.

Answer 2

Looking at the graphic of the functions $$\begin{cases}h(x)=\cos(f(x))+ \dfrac{8}{\pi^3}(f(x))^3 +2f(x)\\g(x)=1+x\end{cases}$$ we verify that $$x=0\Rightarrow h(x)=1\Rightarrow f(0)=0\\x=\pi\Rightarrow h(x)=1+\pi\Rightarrow f(\pi)=\frac{\pi}{2}$$ On the other hand $$f'(x)=\frac{1}{-\sin(f(x)+\dfrac{24f^2(x)}{\pi^3}+2}$$ It follows $f'(0)=\dfrac 12$ and $f'(\pi)=\dfrac{\pi}{6+\pi}$ from which the answer.

Answer 3

Define $g(x) = \cos x + \frac 8{\pi^3}x^3 + 2x$. Since $g'(x) = \frac{24}{\pi^3} x^2 + 2 - \sin x > 0 $, $g$ is strictly increasing, continuous and unbounded on $\mathbb R$. Thus, it is bijective. We can conclude that $g(f(x)) = 1 + x$ not only has a unique solution $f$, it is differentiable and bijective on $\mathbb R$. Its derivative is given by $$f'(x) = \frac 1{g'(f(x))}.$$

Thus, we need to find out $f(0)$ and $f(\pi)$. We could just as well guess the values. Let $x_0, x_1$ be such that $f(x_0) = 0$, $f(x_1) = \pi/2$. We have $$1+x_0 = g(f(x_0)) = g(0) = 1\implies x_0 = 0, \\ 1+x_1 = g(f(x_1)) = g(\pi/2) = 1 + \pi \implies x_1 = \pi. $$

Notice that it is important that we know that such $x_0$ and $x_1$ exist in the first place by surjectivity of $f$. It now follows that $$2f'(0)+f'(\pi) = \frac 2{g'(0)} + \frac 1{g'(\pi/2)} = 1 + \frac \pi{\pi+6}.$$