College Algebra, Chapter 3, 3.7, Section 3.7, Problem 82

Suppose that a car dealership advertises a $\$ 15 \%$ discount on all its new cars. In addition, the manufacturer offers a $\$1000$ rebate on the purchase of a new car. Let $x$ represent the sticker price of the car.

a.) Suppose only the $15 \%$ discount applies. Find a function $f$ that models the purchase price of the car as a function of the price $x$.

b.) Suppose only the $\$ 1000$ rebate applies. Find a function $g$ that models the purchase price of the car as a function of the price $x$.

c.) Find a formula for $H = f \circ g$.

d.) Find $H^{-1}$. What does $H^{-1}$ represent?

e.) Find $H^{-1} (13, 000)$. What does your answer represent?



a.) The purchase price of the car is equal to the difference of the regular price of the car and the discounted price. Thus,

$f(x) = x - 0.15x = 0.85x$

b.) If only $\$ 1000$ rebate applies, then

$g(x) = x - 1000$

c.)


$
\begin{equation}
\begin{aligned}

H = f \circ g = f(g(x)) =& f(x - 1000) = 0.85 (x - 1000)
\\
\\
H(x) =& 0.85 (x - 1000)

\end{aligned}
\end{equation}
$


d.) To find for $H^{-1}$, we set $y = H(x)$


$
\begin{equation}
\begin{aligned}

y =& 0.85(x - 1000)
&& \text{Solve for $x$; divide both sides by } 0.85
\\
\\
\frac{y}{0.85} =& x - 1000
&& \text{Add } 1000
\\
\\
x =& \frac{y}{0.85} + 1000
&& \text{Interchange $x$ and $y$}
\\
\\
y =& \frac{x}{0.85} + 1000
&&

\end{aligned}
\end{equation}
$


Thus, $\displaystyle H^{-1} (x) = \frac{x}{0.85} + 1000$

If $H(x)$ represents the over all discounted price of the car, then $H^{-1} (x)$ represents the original price of the car, without discounts and rebates.

e.)


$
\begin{equation}
\begin{aligned}

H^{-1} (13,000) =& \frac{13,000}{0.85} + 1000
\\
\\
=& \$ 16,294.12

\end{aligned}
\end{equation}
$


It shows that when the discounted price of the car is $\$ 13,000$, its original price is $\$ 16, 294.12$.