Shawn Peck

2022-10-08

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

I'm not sure at all if I did it right but I have taken the function:

$f(x)=x$

Then take the given intervals and insert for $x$ the beginning of interval:

$f(0)=0$

And the end of the interval:

$f(1)=1>0$

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 ${x}_{1}=[0,1]$.

Did I do it correctly? Did I explain correctly?

I'm not sure at all if I did it right but I have taken the function:

$f(x)=x$

Then take the given intervals and insert for $x$ the beginning of interval:

$f(0)=0$

And the end of the interval:

$f(1)=1>0$

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 ${x}_{1}=[0,1]$.

Did I do it correctly? Did I explain correctly?

Derick Ortiz

Beginner2022-10-09Added 11 answers

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

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