Single Variable Calculus, Chapter 1, 1.3, Section 1.3, Problem 32

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


$f(x) = x-2 , \qquad g(x) = x^2 + 3x + 4$


$\begin{equation} \begin{aligned} \text{(a)} \qquad \quad f \circ g &= f(g(x)) && \text{ Substitute the given function $g(x)$ to the value of $x$ of the function $f(x)$}\\ f(x^2+3x+4) &= x-2 && \text{ Simplify the equation}\\ f(x^2+3x+4)&= x^2+3x+4-2 && \text{ Combine like terms} \end{aligned} \end{equation}$


$\boxed{f \circ g =x^2+3x+2}$


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



$\begin{equation} \begin{aligned} \text{(b)} \qquad \quad g \circ f &= g(f(x))\\ g(x-2) &= x^2+3x+4 && \text{ Substitute the given function $g(x)$ to the value of $x$ of the function $f(x)$}\\ g(x-2)&= (x-2)^2+3(x-2)+4 && \text{ Simplify the equation}\\ g(x-2) &= x^2 - 4x+4+3x-6+4 && \text{ Combine like terms} \end{aligned} \end{equation}$



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


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


$\begin{equation} \begin{aligned} \text{(c)} \qquad \quad f \circ f &= f(f(x)) \\ f(x-2) &= x-2 && \text{ Simplify the equation}\\ f(x-2) &= x-2-2 && \text{ Combine like terms} \end{aligned} \end{equation}$


$\boxed{f \circ f=x-4}$


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



$\begin{equation} \begin{aligned} \text{(d)} \qquad \quad g \circ g &= g(g(x)) \\ g(x^2+3x+4) &= x^2+3x+4 && \text{ Substitute the given function $g(x)$ to the value of $x$ of the function $f(x)$}\\ g(x^2+3x+4) &= (x^2+3x+4)^2+3(x^2+3x+4)+4 && \text{ Simplify the equation}\\ g(x^2+3x+4) &= x^4+3x^3+4x^2+3x^3+9x^2+12x+4x^2+12x+16+3x^2+9x+12+4 && \text{ Combine like terms} \end{aligned} \end{equation}$



$\boxed{g \circ g= x^4+6x^3+20x^2+33x+32}$


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