Question

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

Answer 1

Let be the function `f(x)=root(3)x-1+x`

This function is continous in `RR` even in `[0,1]`

Now evaluate the function in extreme points

`f(0)=0-1+0=-1<0`
`f(1)=1-1+1=1>0`

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

And this is what we want to proof. QED