College Algebra, Chapter 3, 3.1, Section 3.1, Problem 24

Evaluate the function $f(x) = x^3 - 4x^2$ at $f(0), \quad f(1), \quad f(-1), f\left( \frac{3}{2} \right), \quad f\left( \frac{x}{2} \right), \quad f(x^2)$ For $f(0)$

$
\begin{equation}
\begin{aligned}
f(0) &= (0)^3 - 4(0)^2 && \text{Replace } x \text{ by } 0\\
\\
&= 0
\end{aligned}
\end{equation}
$


For $f(1)$

$
\begin{equation}
\begin{aligned}
f(1) &= (1)^3 - 4(1)^2 && \text{Replace } x \text{ by } 1\\
\\
&= 1-4 && \text{Simplify}\\
\\
&= -3
\end{aligned}
\end{equation}
$


For $f(-1)$

$
\begin{equation}
\begin{aligned}
f(-1) &= (-1)^3 - 4(-1)^2 && \text{Replace } x \text{ by } -1\\
\\
&= -1-4 && \text{Simplify}\\
\\
&= -5
\end{aligned}
\end{equation}
$


For $f\left( \frac{3}{2} \right)$

$
\begin{equation}
\begin{aligned}
f\left( \frac{3}{2} \right) &= \left( \frac{3}{2} \right)^3 - 4\left( \frac{3}{2} \right)^2 && \text{Replace } x \text{ by } \frac{3}{2}\\
\\
&= \frac{27}{8} - \cancel{4}\left( \frac{9}{\cancel{4}} \right) && \text{Simplify}\\
\\
&= \frac{27}{8} - 9 && \text{Get the LCD}\\
\\
&= \frac{27-72}{8} && \text{Simplify}\\
\\
&= \frac{-45}{8}
\end{aligned}
\end{equation}
$



For $f\left( \frac{x}{2} \right)$

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

For $f(x^2)$

$
\begin{equation}
\begin{aligned}
f(x^2) &= (x^2)^3 - 4(x^2)^2 && \text{Replace } x \text{ by } x^2\\
\\
&= x^6 - 4x^4
\end{aligned}
\end{equation}
$

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