Single Variable Calculus, Chapter 1, 1.3, Section 1.3, Problem 61

(a) Suppose you have $g(x) = 2x+1$ and $h(x) = 4x^2 + 4x +7$, find a function $f$ such that $f o g = h$.

In order to obtain the equation of $f(x)$, we must consider first that the highest degree of the function $h$ is 2 and the highest degree of $g$ is 1, then the function $f(x$) must have the highest degree of 2 also.

Let $f(x) = Ax^2 + Bx + C$ where $A, B, C$ are constant


$
\begin{equation}
\begin{aligned}

f \circ g(x) &= f(g(x))\\
f(2x + 1) &= Ax^2 + Bx + C\\
f(2x + 1) &= A(2x + 1)^2 + B(2x + 1)+ C\\
f \circ g & = 4Ax^2 + 4Ax + A + 2Bx + B + C\\

\end{aligned}
\end{equation}
$


To obtain $f \circ g = h$

$f \circ g = h$


$
\begin{equation}
\begin{aligned}

4Ax^2 + 4Ax + A + 2Bx + B + C &= 4x^2 + 4x + 7\\

\end{aligned}
\end{equation}
$


To solve the value of $A$


$
\begin{equation}
\begin{aligned}

\frac{\cancel{4}A\cancel{x^2}}{\cancel{4}\cancel{x^2}} &= \frac{\cancel{4x^2}}{\cancel{4x^2}}\\
A &= 1

\end{aligned}
\end{equation}
$


To solve the value of $B$


$
\begin{equation}
\begin{aligned}

4Ax + 2Bx &= 4x\\
4(1)x + 2Bx &= 4x\\
4x + 2Bx &= 4x\\
2Bx &= 4x - 4x\\
\frac{\cancel{2}B\cancel{x}}{\cancel{2}\cancel{x}} &= \frac{0}{2x}\\
B &= 0

\end{aligned}
\end{equation}
$


To solve the value of $C$


$
\begin{equation}
\begin{aligned}

A + B + C &= 7\\
1 + 0 + C &= 7\\
C &= 7 - 1\\
C &= 6

\end{aligned}
\end{equation}
$


Substitute values of $A, B$ & $C$ to the function $f(x) = Ax^2 + Bx + C$

$f(x) = (1)(x^2) + (0)(x) + 6$

$\fbox{$f(x) = x^2 + 6$}$

(b) Assume that $f(x) = 3x + 5$ and $h(x) = 3x^2 + 3x + 2$, find a function $g$ such that $f \circ g = h$
In order to obtain function $g(x)$, we must know that the highest degree of $h$ is 2 and the higest degree of $f$ is 1, then $g$ must have the highest degree which is 2.
Let $g(x) = Ax^2 + Bx + C$ where $A, B,$ and $C$ are constant then


$
\begin{equation}
\begin{aligned}

f \circ g (x) =& f(g(x))\\
f(Ax^2 + Bx + C) =& 3x + 5\\
f \circ g =& 3Ax^2 + 3Bx + 3C + 5\\
\text { To obtain } f \circ g =& h\\
3Ax^2 + 3Bx + 3C + 5 =& 3x^2 + 3x + 2\\
\end{aligned}
\end{equation}
$


Solving for the value of $A, B, C$


$
\begin{equation}
\begin{aligned}

\frac{\cancel{3} A \cancel{x^2}}{\cancel{3}\cancel{x^2}} =& \frac{\cancel{3x^2}}{\cancel{3x^2}}\\
A =& 1\\
\\
\frac{\cancel{3}B \cancel{x}}{\cancel{3x}} =& \frac{\cancel{3x}}{\cancel{3x}}\\
B =& 1\\
\\
3C + 5 =& 2\\
3C =& 2-5\\
\frac{3C}{3} =& \frac{-3}{3}\\
C =& -1
\end{aligned}
\end{equation}
$


Substitute the value $A, B, C$ to function $g(x) = Ax^2 + Bx + C$

$g(x) = (1)(x^2) + (1)(x) + (-1)$
$\fbox{$g(x) = x^2 + x - 1$}$