Single Variable Calculus, Chapter 1, 1.3, Section 1.3, Problem 62

Suppose $f(x) = x+4$ and $h(x)=4x-1$, find $a$ function $g$ such that $g \circ f = h$.
Both equations are linear functions so function $g(x)$ must be a linear function as well.
Let $g(x) = Ax + B$, where $A$ and $B$ are constant.



$
\begin{equation}
\begin{aligned}

g \circ f(x) =& g(f(x))\\
g(x + 4) =& Ax + B\\
g(x + 4) =& A(x + 4) + B\\
g \circ f =& Ax + 4A + B


\end{aligned}
\end{equation}
$


To obtain $g \circ f = h$

$Ax + 4A + B = 4x - 1$
By grouping the equation in terms of the linear equation $Ax + B$


$
\begin{equation}
\begin{aligned}

\frac{Ax}{x} =& \frac{4x}{x}\\
A =& 4\\
\\
4A + B =& -1\\
4(4) + B =& -1\\
16 + B =& -1\\
B =& -17

\end{aligned}
\end{equation}
$


Substituting to the function $g(x) = Ax + B$

$g(x) = (4)(x) + (-17)$
$\fbox{$g(x) = 4x - 17$}$