0

Let the function $f(x)$ be thrice differentiable and satisfies $f(f(x))=1-x$ for all $x\in[0,1]$. Then find the value of $$\int_0^1f(x)dx$$ given that $f''(\frac{4}{5})=0$.

MY APPROACH:

I differentiated the function twice to get $f''(f(x))(f'(x))^2+f''(x)f'(f(x))=0$. Now I substituted $x=\frac{4}{5}$.

Which led to $f''(f(\frac{4}{5})(f'(\frac{4}{5}))^2=0$ and $f'(\frac{4}{5})\ne 0$ from the first derivative. i.e. I get $f(\frac{4}{5})=\frac{4}{5}$ i.e. $f(x)=x$. Am I right?

Please help. THANKS

Savitar
  • 174

1 Answers1

2

I don't think such function exists since $f$ is injective because if $f(x)=f(y)$ then $1-x=f(f(x))=f(f(y))=1-y$ implies $x=y$. Also we have that $f(f(x))$ is decreasing clearly.

And since $f$ is differentiable it is continuous, so it's either strictly increasing or decreasing.

  • If $f$ is strictly decreasing then $f\circ f$ is strictly increasing which is a contradiction
  • If $f$ is strictly increasing then $f\circ f$ is strictly increasing which is a contradiction.

If the function weren't differentiable then since $f(f(x))=1-x$ then $f(1-x)=1-f(x)$ so $$\int_0^1 f(x)dx=1-\int_0^1f(1-x)dx=1-\int_0^1f(x)\implies\int_0^1f(x)=\frac12$$

kingW3
  • 13,496