If \( f(x) = x^2 \) and \( g(x) = 2^x \)
(a) Find \( g(f(2)) \)
(b) What is the range of \( g(f(x)) \) ?
(c) Show \( g(-x) = 1/g(x) \).
Substitute the function definitions from the outside in.
\[ \begin{align}
g(f(x)) &= 2^{f(x)} \\
&= 2^{x^2}
\end{align} \ \]
so
\[ \begin{align}
g(f(2)) &= 2^{2^2} \\
&= 2^{4} \\
&= 16
\end{align} \ \]
The best way to determine the range is to substiute extreme x-values.
Both functions are defined for all x. However, neither function takes y-values less than zero. So the range is y values greater than 0.
To show that g(-x) = 1/g(x), substitute -x in the definition
\[ \begin{align}
g(-x) &= 2^{-x} \\
&= \frac{1}{2^x} \\
&= \frac{1}{g(x)} \\
\end{align} \ \]