# Use the intermediate value theorem to show that there is a root of the equation x^5-2x^4-x-3=0 in the interval (2,3)?

##### 1 Answer
May 23, 2018

See below for proof.

#### Explanation:

If $f \left(x\right) = {x}^{5} - 2 {x}^{4} - x - 3$
then
$\textcolor{w h i t e}{\text{XXX}} f \left(\textcolor{b l u e}{2}\right) = {\textcolor{b l u e}{2}}^{5} - 2 \cdot {\textcolor{b l u e}{2}}^{4} - \textcolor{b l u e}{2} - 3 = \textcolor{red}{- 5}$
and
$\textcolor{w h i t e}{\text{XXX}} f \left(\textcolor{b l u e}{3}\right) = {\textcolor{b l u e}{3}}^{5} - 2 \cdot {\textcolor{b l u e}{3}}^{4} - \textcolor{b l u e}{3} - 3 = 243 - 162 - 3 - 3 = \textcolor{red}{+ 75}$

Since $f \left(x\right)$ is a standard polynomial function, it is continuous.

Therefore, based on the intermediate value theorem, for any value, $\textcolor{m a \ge n t a}{k}$, between $\textcolor{red}{- 5}$ and $\textcolor{red}{+ 75}$, there exists some $\textcolor{\lim e}{\hat{x}}$ between $\textcolor{b l u e}{2}$ and $\textcolor{b l u e}{3}$ for which $f \left(\textcolor{\lim e}{\hat{x}}\right) = \textcolor{m a \ge n t a}{k}$

Since $\textcolor{m a \ge n t a}{0}$ is such a value,
there exists some value $\textcolor{\lim e}{\hat{x}} \in \left[\textcolor{b l u e}{2} , \textcolor{b l u e}{3}\right]$ such that $f \left(\textcolor{\lim e}{\hat{x}}\right) = \textcolor{m a \ge n t a}{0}$