Calculus - Interval Bisection Method

Question

For x^5-x=3, explain why there must be a zero of the function between (0,2).

I know that I should use the intermediate value theorem to prove this, but I can't think of a good way to explain this.

Answer 1

$f(x)=x^5-x-3$, $f(0)=-3$, $f(2)=27$. By IVT, there is a zero between $0$ and $2$ if $f$ is continuous which it is.