2

$f(f(x))=x \,\forall x\in [0,1], f(0)=1$and $f$ is differentiable.Find $f$.

I can guess two such

$f(x)=1-x, f(x)=\frac {1-x}{ax+1}$ where $a > -1$ is some real.

I got one general solution for $f(f(x))=x$.

Suppose $h(x)$ is one such function and g be any invertible function then $f(x)=g(h(g^{-1}(x)))$

Makar
  • 2,267
  • So, what's your question? –  Dec 11 '20 at 08:02
  • I think $a$ need not be positive. $a>-1$ is good enough. – Kavi Rama Murthy Dec 11 '20 at 08:08
  • 2
    The given condition simply says that $f$ is its own inverse function on $[0,1]$, which means only that its graph must be invariant under reflection in the line $y=x$. There are bazillions of such functions—just draw any curve from $(0,1)$ to the line $y=x$ that is differentiable, decreasing, stays positive, and meets the line $y=x$ at a right angle; then it together with its reflection form such a function. – Greg Martin Dec 11 '20 at 08:18

1 Answers1

1

Hinnt:

$f(x)=\frac{ax+b}{cx-a}, x\ne c/a$ follows $f(f(x))=x$

$f(x)=(1-x^k)^{1/k}$ follows $f(f(x))=x$.

Z Ahmed
  • 43,235