Single Variable Calculus, Chapter 2, 2.5, Section 2.5, Problem 48

Use the Intermediate Value Theorem to show that $\sqrt[3]{x} = 1 - x$ has root on the interval $(0, 1)$

Let $f(x) = \sqrt[3]{x}+x-1$
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 0 and 1 such that $f(x) = 0$ and that is, $f(c) = 0 $.


$
\begin{equation}
\begin{aligned}

f(0) =& \sqrt[3]{0} + 0 -1 = -1\\
\\
f(1) =& \sqrt[3]{1} + 1 - 1 = 1

\end{aligned}
\end{equation}
$


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


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

\end{aligned}
\end{equation}
$

Therefore,

There exist a such solution $c$ for $\sqrt[3]{x} + x - 1 = 0$