Let $f(x)$ be a four-times differential function such that
$f(2x^2-1)=2xf(x)$
What is the value of $f''''(0)$?
A brute force approach is differentiating the given condition 4 times and finding values of $f(x)$ and its derivatives at required numbers or at a value of $x$ where the function can be simplified. Considering the LHS has a composite function and the RHS has the product of 2 functions, differentiating both sides leads to a lot of terms. I tried shifting values of $x$ but did not find a convenient substitution to simplify the problem. How should I proceed?
Best Answer
I managed to figure this out a little late.
In the given functional equation, replace $x$ with $-x$.
The equation is now written as:
$f(2x^2-1)=-2xf(-x)$
But it is given that $f(2x^2-1)=2xf(x)$
Therefore, $-2xf(-x)=2xf(x)$ $=>$ $-f(-x)=f(x)$
A rather simple functional equation which can be easily differentiated 4 times.
The answer is $0$.