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}$


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