Single Variable Calculus, Chapter 1, 1.3, Section 1.3, Problem 31

We need to find (a) $f \circ g$, (b) $g \circ f$, (c) $f \circ f$, and (d) $g \circ g$ and state their domains


$f(x) = x^2 - 1, \qquad g(x) 2x + 1$




$
\begin{equation}
\begin{aligned}
\text{(a)}     f \circ g &= f(g(x))\\
f(2x+1) &= x^2-1 && \text{ Substitute the given values of the function $f(x)$ and $g(x)$ }\\
f(2x+1) &= (2x+1)^2-1 && \text{ Simplify the equation}\\
f(2x+1) &= 4x^2+4x+1-1 && \text{ Combine like terms}\\
f(2x+1) &= 4x^2+4x

\end{aligned}
\end{equation}
$


$\boxed{\text{ The domain of this function is}   (-\infty,\infty)} $



$
\begin{equation}
\begin{aligned}

\text{(b)}     g \circ f =& g(f(x))\\
g(x^2-1)=& 2x+1 && \text{ Substitute the given function $g(x$) to the value of $x$ of the function $f(x)$}\\
g(x^2-1)=& 2(x^2-1)+1 && \text{ Simplify the equation}\\
g(x^2-1)=& 2x^2-2+1 && \text{ Combine like terms}

\end{aligned}
\end{equation}
$


$ \boxed{g \circ f=2x^2-1}$


$ \boxed{ \text{ The domain of this function is }   (-\infty,\infty)}$


$
\begin{equation}
\begin{aligned}
\text{(c)}     f \circ f &= f(f(x))\\
f(x^2-1) &= x^2-1 && \text{ Substitute the given function $g(x$) to the value of $x$ of the function $f(x)$}\\
f(x^2-1) &= (x^2-1)^2-1 && \text{ Simplify the equation}\\
f(x^2-1) &= x^4-x^2-x^2+1-1 && \text{ Combine like terms}

\end{aligned}
\end{equation}
$


$\boxed{f \circ f=x^4-2x^2}$
$\boxed{\text{ The domain of this function is } (-\infty,\infty)}$



$
\begin{equation}
\begin{aligned}

\text{(d)}     g \circ g &= g(g(x))\\
g(2x+1) &= 2x+1 && \text{ Substitute the given function $g(x$) to the value of $x$ of the function $f(x)$:}\\
g(2x+1) &= 2(2x+1)+1 && \text{ Simplify the equation}\\
g(2x+1) &= 4x+2+1 && \text{ Combine like terms}\\

\end{aligned}
\end{equation}
$


$\boxed{g \circ g =4x+3}$
$\boxed{\text{ The domain of this function is } (-\infty,\infty)}$