Exercise 1.3.67

Question

\begin{enumerate}[label=(\alph*)]
\item    If $g(x)=2 x+1$ and $h(x)=4 x^{2}+4 x+7$, find a funcion $f$ such that $f \circ g=h$. (Think about what operations you would have to perform on the formula for $g$ to end up with the formula for $h$.)
\item If $f(x)=3 x+5$ and $h(x)=3 x^{2}+3 x+2$, find a function $g$ such that $f \circ g=h$.
\end{enumerate}

Khagan Gulusoy · Accepted Answer

(a)

$$
\begin{aligned}
& f \circ g=h \quad f(2 x+1)=4 x^{2}+4 x+7 . \
& 4 x^{2}+4 x+7=(2 x+1)^{2}+6  \implies \forall x \in \mathbb{R}: f(x)=x^{2}+6 .
\end{aligned}
$$

(b)

$$
\begin{aligned}
f(g(x))=h \implies \quad & \forall x \in \mathbb{R}: \quad 3 g(x)+5=3 x^{2}+3 x+2 . \
& \forall x \in \mathbb{R}: \quad 3 g(x)=3 x^{2}+3 x-3 \
& \forall x \in \mathbb{R}: \quad g(x)=x^{2}+x-1 .
\end{aligned}
$$

Exercise 1.3.67

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Exercise 1.3.67

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