# How do you use the intermediate value theorem to verify that there is a zero in the interval [0,1] for f(x)=x^3+x-1?

Aug 14, 2016

There is exactly 1 zero in this interval.

#### Explanation:

The intermediate value theorem states that for a continuous function defined on interval $\left[a , b\right]$ we can let $c$ be a number with
$f \left(a\right) < c < f \left(b\right)$ and that $\exists x \in \left[a , b\right]$ such that $f \left(x\right) = c$.

A corollary of this is that if the sign of $f \left(a\right) \ne$ sign of $f \left(b\right)$ this means that there must be some $x \in \left[a , b\right]$ such that $f \left(x\right) = 0$ because $0$ is obviously between the negatives and positives.

So, let's sub in the endpoints:

$f \left(0\right) = {0}^{3} + 0 - 1 = - 1$

$f \left(1\right) = {1}^{3} + 1 - 1 = 1$

$\therefore$ there is at least one zero in this interval. To check if there's only one root we look at the derivative which gives the slope.

$f ' \left(x\right) = 3 {x}^{2} + 1$

We can see that $\forall x \in \left[a , b\right] , f ' \left(x\right) > 0$ so the function is always increasing in this interval - this means there is only one root in this interval.