# How do you use the intermediate value theorem to explain why f(x)=x^3+3x-2 has a zero in the interval [0,1]?

Nov 19, 2016

$f \left(x\right)$ is continuous in the domain $x \in \left[0 , 1\right]$ and
$f \left(0\right) < 0 < f \left(1\right)$
therefore ${\exists}_{c} : f \left(c\right) = 0$ (by the intermediate value theorem)

#### Explanation:

The intermediate value says:

If $f \left(x\right)$ is a continuous function over the domain $x \in \left[a , b\right]$
then for any value $k \in \left[f \left(a\right) , f \left(b\right)\right]$
there is a value $c$ such that $f \left(c\right) = k$

For the given function $f \left(0\right] = - 2$ and $f \left(1\right) = + 2$
Since $k = 0 \in \left[- 2 , + 2\right]$
there is a value $c$ such that $f \left(c\right) = 0$

[Technically, I have assumed without proof that $f \left(x\right)$ is continuous within the given domain. If you need this proof, add as a new question or post a comment.]