College Algebra, Chapter 3, 3.1, Section 3.1, Problem 22

Evaluate the function $h(t) = t + \frac{1}{t}$ at $h(1), \quad h(-1), \quad h(2), h\left( \frac{1}{2} \right), \quad h(x), h\left( \frac{1}{x} \right)$
For $h(1)$

$
\begin{equation}
\begin{aligned}
h(1) &= 1 + \frac{1}{1} && \text{Replace } t \text{ by } 1\\
\\
&= 1 + 1 && \text{Simplify}\\
\\
&= 2
\end{aligned}
\end{equation}
$


For $h(-1)$

$
\begin{equation}
\begin{aligned}
h(-1) &= 1 + \frac{1}{-1} && \text{Replace } t \text{ by } -1\\
\\
&= -1 - 1 && \text{Simplify}\\
\\
&= -2
\end{aligned}
\end{equation}
$


For $h(2)$

$
\begin{equation}
\begin{aligned}
h(2) &= 2 + \frac{1}{2} && \text{Replace } t \text{ by } 2\\
\\
&= \frac{4+1}{2} && \text{Get the LCD}\\
\\
&= \frac{5}{2}
\end{aligned}
\end{equation}
$


For $h\left( \frac{1}{2} \right)$

$
\begin{equation}
\begin{aligned}
h\left( \frac{1}{2} \right) &= \frac{1}{2} + \frac{1}{\frac{1}{2}} && \text{Replace } t \text{ by } \frac{1}{2}\\
\\
&= \frac{1}{2} + 2 && \text{Get the LCD}\\
\\
&= \frac{1+4}{2} && \text{Simplify}\\
\\
&= \frac{5}{2}
\end{aligned}
\end{equation}
$


For $h(x)$

$
\begin{equation}
\begin{aligned}
h(x) &= x + \frac{1}{x} && \text{Replace } t \text{ by } x\\
\end{aligned}
\end{equation}
$


For $h\left( \frac{1}{x} \right)$

$
\begin{equation}
\begin{aligned}
h\left( \frac{1}{x} \right) &= \frac{1}{x} + \frac{1}{\frac{1}{x}} && \text{Replace } t \text{ by } \frac{1}{x}\\
\\
&= \frac{1}{x} + x
\end{aligned}
\end{equation}
$