# How do you use the intermediate value theorem to prove that f(x)=x^3-9x+5 has a real zero in the intervals [-4,-3], [0,1], and [2,3]?

Aug 23, 2017

See explanation...

#### Explanation:

Given:

$f \left(x\right) = {x}^{3} - 9 x + 5$

By the intermediate value theorem (or by the simpler Bolzano's theorem), the function $f \left(x\right)$ has a zero in an interval $\left[a , b\right]$ whenever $f \left(a\right)$ and $f \left(b\right)$ have opposite signs. This is a sufficient but not necessary condition.

With the given $f \left(x\right)$ we find:

$\left\{\begin{matrix}f \left(- 4\right) = {\left(\textcolor{b l u e}{- 4}\right)}^{3} - 9 \left(\textcolor{b l u e}{- 4}\right) + 5 = - 64 + 36 + 5 = - 23 < 0 \\ f \left(- 3\right) = {\left(\textcolor{b l u e}{- 3}\right)}^{3} - 9 \left(\textcolor{b l u e}{- 3}\right) + 5 = - 27 + 27 + 5 = 5 > 0\end{matrix}\right.$

So $f \left(x\right)$ has a zero in $\left[- 4 , - 3\right]$

$\left\{\begin{matrix}f \left(0\right) = {\left(\textcolor{b l u e}{0}\right)}^{3} - 9 \left(\textcolor{b l u e}{0}\right) + 5 = 5 > 0 \\ f \left(1\right) = {\left(\textcolor{b l u e}{1}\right)}^{3} - 9 \left(\textcolor{b l u e}{1}\right) + 5 = 1 - 9 + 5 = - 3 < 0\end{matrix}\right.$

So $f \left(x\right)$ has a zero in $\left[0 , 1\right]$

$\left\{\begin{matrix}f \left(2\right) = {\left(\textcolor{b l u e}{2}\right)}^{3} - 9 \left(\textcolor{b l u e}{2}\right) + 5 = 8 - 18 + 5 = - 5 < 0 \\ f \left(3\right) = {\left(\textcolor{b l u e}{3}\right)}^{3} - 9 \left(\textcolor{b l u e}{3}\right) + 5 = 27 - 27 + 5 = 5 > 0\end{matrix}\right.$

So $f \left(x\right)$ has a zero in $\left[2 , 3\right]$

graph{x^3-9x+5 [-7, 7, -10, 20]}

$\textcolor{w h i t e}{}$
Notes

The intermediate value theorem is one way of expressing the completeness of the real numbers. It basically says that there are no gaps in the real number line that would allow a continuous function to change sign without being equal to zero at some point.

The same is not true of the rational numbers. For example, the function:

$f \left(x\right) = {x}^{2} - 2$

has the properties:

$\left\{\begin{matrix}f \left(1\right) = - 1 < 0 \\ f \left(2\right) = 2 > 0\end{matrix}\right.$

but there is no rational number $x$ such that $f \left(x\right) = 0$.

So there are "gaps" in the rational numbers - in this particular case the number we would call $\sqrt{2}$.