If
\[ \begin{align}
f(x) &= \frac{1}{\sqrt x} \quad \text{and} \quad g(x)= x^2-1
\end{align}\ \]
(a) Find f(g(x))
(b) Find g(f(x))
(c) What is the domain of f(g(x))?
Substitute from the outside in
\[ \begin{align}
f(g(x)) &= \frac{1}{\sqrt{g(x)}} \\
&= \frac{1}{\sqrt {x^2 - 1}}
\end{align}\ \]
Substitute from the outside in
\[ \begin{align}
g(f(x)) &= (f(x))^2 - 1 \\
&= (\frac{1}{\sqrt x})^2 - 1 \\
&= \frac{1}{x} - 1
\end{align}\ \]
The domain of f(g(x)) is all x except those points where f(x) is not defined and g(x) is not defined
f(x) is not defined when g(x) = 0
\[ \begin{align}
x^2 - 1 &= 0 \\
(x-1)(x+1) &= 0 \\
x &= \pm 1
\end{align}\ \]
g(x) is not defined when \(x^2-1\) is less than 0.
\[ \begin{align}
x^2 - 1 &\lt 0 \\
x^2 &\lt 1 \\
-1 \lt x &\lt 1 \\
\end{align}\ \]
Combining the two sets of points, the domain of f(g(x)) is all x except \(-1 \le x \le 1 \).