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 $f (0.2)$,

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