Single Variable Calculus, Chapter 2, 2.5, Section 2.5, Problem 47

Use the Intermediate Value Theorem to show that $x^4 + x - 3 = 0$ has root on the interval $(1, 2)$

Let $f(x) = x^4 + x - 3$

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 that
given interval. So, there exist a number $c$ between 1 and 2 such that $f(x) = 0$ and that is, $f(c) = 0 $.


$
\begin{equation}
\begin{aligned}

f(1) =& (1)^4 + 1 - 3 = -1\\
\\
f(2) =& (2)^4 + 2 - 3 = 15

\end{aligned}
\end{equation}
$

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


So,
$
\begin{equation}
\begin{aligned}
& \text{ if } 1 < c < 2 \quad \text{ then } \quad f(1) < f(c) < f(2)\\
& \text{ if } 1 < c < 2 \quad \text{ then } \quad -1 < 0 < 15

\end{aligned}
\end{equation}
$



Therefore,

There exist a such solution $c$ for $x^4+x-3=0$