# Use the intermediate Value Theorem to show that x^(1/3)=1-x have at least a solution in [0,1]?

May 25, 2018

Let be the function $f \left(x\right) = \sqrt[3]{x} - 1 + x$

This function is continous in $\mathbb{R}$ even in $\left[0 , 1\right]$

Now evaluate the function in extreme points

$f \left(0\right) = 0 - 1 + 0 = - 1 < 0$
$f \left(1\right) = 1 - 1 + 1 = 1 > 0$

The intermediate value theorem (based on Bolzano's theorem) establish that exist a number $c \in \left[0 , 1\right]$ such that $f \left(c\right) = 0$

And this is what we want to proof. QED