Single Variable Calculus, Chapter 2, 2.5, Section 2.5, Problem 52

(a) Show that the equation $x^5 - x^2 + 2x + 3 = 0$ has at least one real root.
(b) Determine an inteval of length 0.01 that contains a root using a calculator.

(a) Let $f(x) = x^5 - x^2 + 2x + 3$
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 the given interval. So we take $a$ and $b$ to be -1 and 1 respectively and assume the function $f(x)$
is continuous on the interval (-1,1). So we have,


$\begin{equation} \begin{aligned} f(-1) =& (-1)^5 - (-1)^2 + 2 (-1) + 3 = -1\\ \\ f(1) =& (1)^5 - (1)^5 - (1)^2 + 2(1) + 3 = 5 \end{aligned} \end{equation}$

By using Intermediate Value Theorem. We prove that...

So,
$\begin{equation} \begin{aligned} & \text{if } -1 < c < 1 && \text{then } \quad f(-1) < f(c) < f(1)\\ & \text{if } -1 < c < 1 && \text{then } \quad -1 < 0 < 5 \end{aligned} \end{equation}$

Therefore,
There exist such root for $x^5 - x^2 + 2x + 3 = 0$.


(b) By trial and error using calculator, we take the interval (-0.88,-0.87) so,


$\begin{equation} \begin{aligned} f(-0.88) =& (-0.88)^5 - (-0.88)^2 + 2(-0.88) + 3 = -0.062 < 0 \\ & \text{ and }\\ f(-0.87) =& (-0.87)^5 - (-0.87)^2 + 2(-0.87) + 3 = 0.005 > 0 \end{aligned} \end{equation}$


Therefore,
The root in the function $x^5 - x^2 + 2x + 3 = 0$ exists between (-0.88,-0.87).