College Algebra, Chapter 3, 3.1, Section 3.1, Problem 18

Evaluate the function $f(x) = x^3 + 2x$ at $f(-2), \quad f(1), \quad f(0), \quad f\left( \frac{1}{3} \right), \quad f(0.2)$
For $f(-2)$,

$
\begin{equation}
\begin{aligned}
f(-2) &= (-2)^3 + 2(-2) && \text{Replace } x \text{ by } -2\\
\\
&= -8-4 && \text{Simplify}\\
\\
&= -12
\end{aligned}
\end{equation}
$


For $f(1)$,

$
\begin{equation}
\begin{aligned}
f(1) &= (1)^3 + 2(1) && \text{Replace } x \text{ by } 1\\
\\
&= 1 + 2 && \text{Simplify}\\
\\
&= 3
\end{aligned}
\end{equation}
$

For $f(0)$,

$
\begin{equation}
\begin{aligned}
f(0) &= (0)^3 + 2(0) && \text{Replace } x \text{ by } 0\\
\\
&= 0
\end{aligned}
\end{equation}
$


For $f\left( \frac{1}{3} \right)$,

$
\begin{equation}
\begin{aligned}
f\left( \frac{1}{3} \right) &= \left( \frac{1}{3} \right)^3 + 2\left( \frac{1}{3} \right) && \text{Replace } x \text{ by } \frac{1}{3}\\
\\
&= \frac{1}{27} + \frac{2}{3} && \text{Get the LCD}
\\
&= \frac{1+11}{27} && \text{Simplify}\\
\\
&= \frac{12}{27} && \text{Reduce to lowest term}\\
\\
&= \frac{4}{9}
\end{aligned}
\end{equation}
$


For $f (0.2)$,

$
\begin{equation}
\begin{aligned}
f(0.2) &= (0.2)^3 + 2 (0.2) && \text{Replace } x \text{ by } 0.2\\
\\
&= 0.008 + 0.4 && \text{Simplify}\\
\\
&= 0.408
\end{aligned}
\end{equation}
$