Single Variable Calculus, Chapter 2, Review Exercises, Section Review Exercises, Problem 27

Use the Intermediate Value Theorem to show that $2x^3 + x^2 + 2 = 0$ has a root in the interval $(-2, -1)$

Let $f(x) = 2x^3 + x^2 + 2$

Based from the definition of Intermediate Value Theorem, there exist a solution $c$ for the function between the interval $(a, b)$. Suppose that the function is continuous on that given interval. So, there exist a
number $c$ between $-2$ and $-1$ such that $f(x) = 0$ and that is, $f(c) = 0$




$
\begin{equation}
\begin{aligned}

\qquad f(-2) =& 2(-2)^3 + (-2)^2 + 2 = -10\\
\\
\qquad f(-1) =& 2(-1)^3 + (-1)^2 + 2 = 1

\end{aligned}
\end{equation}
$


By using Intermediate Value Theorem, we prove that

$\qquad$ if $-2 < c < - 1$ then $f(-2) < f(c) < f(-1)$

So,

$\qquad$ if $-2 < c < -1$ then $-10 < 0 < 1$

Therefore,

$\qquad$ There exist a such solution $c$ for $2x^3 + x^2 + 2 = 0$