College Algebra, Chapter 3, Review Exercises, Section Review Exercises, Problem 76

If $f(x) = 1 + x^2$ and $g(x) = \sqrt{x-1}$, find the following.

$
\begin{equation}
\begin{aligned}
&\text{a.) } f \circ g &&& &\text{b.) } g \circ f\\
\\
&\text{c.) } (f \circ g)(2) &&& &\text{d.) } (f \circ f) (2) \\
\\
&\text{e.) } f \circ g \circ f &&& &\text{f.) } g \circ f \circ g
\end{aligned}
\end{equation}
$


a.) $f \circ g = f(g(x)) = f(\sqrt{x-1}) = 1 + (\sqrt{x-1})^2 = 1 + x - 1 = x$
b.) $g \circ f = g(f(x)) = g(1+x^2) = \sqrt{1+x^2-1} = \sqrt{x^2} = x$
c.) $(f \circ g)(2) = 2$
d.) $(f \circ f)(2)$

$
\begin{equation}
\begin{aligned}
f \circ f = f(f(x)) = f(1 + x^2) &= 1 + (1 + x^2)^2\\
\\
&= 1 + 1 + 2x^2 + x^4\\
\\
&= x^4 + 2x^2 + 2 \\
\\
(f \circ f)(2) &= (2)^4 + 2(2)^2 + 2\\
\\
&= 26
\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}
\text{e.) } f \circ g \circ f &= f(g(f(x)))\\
\\
&= f(g(1+x^2))\\
\\
&= f\left(\sqrt{1+x^2-1}\right)\\
\\
&= f\left(\sqrt{x^2}\right)\\
\\
&= f(x) \\
\\
&= 1 + x^2
\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}
\text{f.) } g \circ f \circ g &= g(f(g(x)))\\
\\
&= g(f(\sqrt{x-1}))\\
\\
&= g\left(1 + (\sqrt{x-1})^2\right)\\
\\
&= g(1 + x -1 )\\
\\
&= g(x) \\
\\
&= \sqrt{x-1}
\end{aligned}
\end{equation}
$