Let $f(x)$ be a fourth differentiable function such that $f(2x^2-1)=2xf(x)$, $\forall x \in \mathbb R$, then $f^{(4)}(0)$ is equal to?
Solution Given in text book:
Replace $x$ by $(-x)$ in given relation
$f(2x^2-1)=2xf(x)$
So in final he obtained $x[f(x)+f(-x)]=0$ and from here deducted that $f(x)$ is an odd function.
My doubt:
How can we surely say that $f(x)+f(-x)=0$?
Can we obtain function from given relation?
Best Answer
There's no doubt that $f(x)$ is an odd function when $x\neq 0$. So let's consider the situation when $x=0$.
Firstly, given that $f(x)$ is a fourth differentiable function, $f(x)$ is necessarily continuous (at every point where it is differentiable). Thus we have the following equality according to the definition: \begin{align*} \lim_{x\to 0+}\frac{f(x)-f(0)}{x}&=\lim_{x\to 0-}\frac{f(x)-f(0)}{x} \\ &=\lim_{-x\to 0+}\frac{f(x)-f(0)}{x}=\lim_{-x\to 0+}\frac{-f(-x)-f(0)}{x}=\lim_{x\to 0+}\frac{-f(x)-f(0)}{-x} \end{align*} $$\lim_{x\to 0+}\frac{f(x)-f(0)}{x}-\frac{-f(x)-f(0)}{-x}=\lim_{x\to 0+}-\frac{2f(0)}{x}=0$$ We get $f(0)=0$