Fixed Point of a Function

Question

I am trying to answer the following question:

Prove the function $f(x)=1-x^2$ has a fixed point on $[0,1]$. Find the value of this fixed point explicitly.

I know how to prove that it has a fixed point using the IVT with $g(x)=f(x)-x$.

$g(0) = 1$

$g(1) = -1$

Thus, it is implied that there is a fixed point.

My question is, how do you find the fixed point explicitly? I am unsure what this even means. Thanks!

Answer 1

Hint: $x_0$ is a fixed point of $f$ if and only if $f(x_0) = x_0$. So you're trying to find an $x_0$ such that

$$x_0 = f(x_0) = 1 - x_0^2$$

Answer 2

To find the fixed point you have to solve $f(x)=x$ which gives us:

$1-x^2=x$, i.e. $1-x-x^2=0$. So, $x=(1+\sqrt5)/2$ (the other root is negative)