Single Variable Calculus, Chapter 2, 2.5, Section 2.5, Problem 53

(a) Show that the equation $x^5 - x^2 - 4 = 0$ has at least one real root.
(b) Determine the root using a graph.

(a) Let $f(x) = x^5 - x^2 - 4$
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 2 respectively and assume the function $f(x)$
is continuous on the interval (1,2). So we have,


$\begin{equation} \begin{aligned} f(1) =& (1)^5 - (1)^2 - 4 = -4\\ \\ f(2) =& (2)^5 - (2)^2 - 4 = 24 \end{aligned} \end{equation}$


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

So,
$\begin{equation} \begin{aligned} & \text{if } 1 < c < 2 && \text{then } \quad f(1) < f(c) < f(2)\\ & \text{if } 1 < c < 2 && \text{then } \quad -4 < 0 < 24 \end{aligned} \end{equation}$

Therefore,
There exist such root for $x^5 - x^2 - 4 = 0$.


(b) Referring to the graph, the approximate value of root is $x = 1.43$