Shawn Peck

2022-10-08

Let $f:\left[0,1\right]\to \mathbb{R}$ continuous and let $f\left(\left[0,1\right]\right)\subset \left[0,1\right]$. By using intermediate value theorem, show that $f$ has at least got one fixed point in $\left[0,1\right]$ (aka there is one solution $x\in \left[0,1\right]$ of the equation $f\left(x\right)=x$).

I'm not sure at all if I did it right but I have taken the function:
$f\left(x\right)=x$
Then take the given intervals and insert for $x$ the beginning of interval:
$f\left(0\right)=0$
And the end of the interval:
$f\left(1\right)=1>0$
Because the function is continuous and because the equality sign changes to an inequality sign right after, the intermediate value theorem provides there must be at least one solution ${x}_{1}=\left[0,1\right]$.
Did I do it correctly? Did I explain correctly?

### Answer & Explanation

Derick Ortiz

Hint: Apply the intermediate value theorem to the function $g\left(x\right)=f\left(x\right)-x$. Note that $g\left(0\right)\ge 0\ge g\left(1\right)$.