Chapter 5.6, Problem 16PPS

a.

To determine

To calculate: the fuel economy when the car is travelling 40 miles per hour.

Answer to Problem 16PPS

The fuel economy is $29.5016$ miles per gallon.

Explanation of Solution

Given information:

A specific car’s fuel economy in miles per gallon can be $f(x)=0.00000056x^4-0.000018x^3-0.016x^2+1.38x-0.38$ where x represents the car’s speed in miles per hour.

Formula used:

Synthetic Substitution :- Synthetic division is used to find the value of a function.

Remainder Theorem:- If a polynomial $P(x)$ is divided by $x-r$ , the remainder s a constant $P(r)$ , and

  $P(x)=Q(x)(x-r)+P(r)$ ,

Where $Q(x)$ is a polynomial with degree one less than $P(x)$ .

Calculation:

Consider ,

  $f(x)=0.00000056x^4-0.000018x^3-0.016x^2+1.38x-0.38$

By using Remainder Theorem, $f(40)$ should be the remainder when the polynomial is divided by $x-40$ .

  $\begin{array}{l}\overline{)40}\begin{array}{cccc}0.00000056& -0.000018& \begin{array}{cc}-0.016& \begin{array}{cc}1.38& -0.38\end{array}\end{array}& \end{array}\\ \begin{array}{cccc}0.0000224& 0.000176& -\text{0}\text{.63296}& \text{ 29}\text{.8816}\end{array}\\ \overline{\begin{array}{cccc}0.00000056& 0.0000044& -0.015824& \text{ 0}\text{.74704}\end{array}\hspace{.2em}|}\text{ 29}\text{.5016}\end{array}$

The remainder is $29.5016$ .

Therefore, by synthetic substitution, $f(40)=\mathrm{29.5016.}$

Hence, the fuel economy when the car is travelling 40 miles per hour is $29.5016$ miles per gallon.

b.

To determine

To calculate: the fuel economy when the car is travelling 50 miles per hour.

Answer to Problem 16PPS

The fuel economy is $29.87$ miles per gallon.

Explanation of Solution

Given information:

A specific car’s fuel economy in miles per gallon can be $f(x)=0.00000056x^4-0.000018x^3-0.016x^2+1.38x-0.38$ where x represents the car’s speed in miles per hour.

Formula used:

Synthetic Substitution :- Synthetic division is used to find the value of a function.

Remainder Theorem:- If a polynomial $P(x)$ is divided by $x-r$ , the remainder s a constant $P(r)$ , and

  $P(x)=Q(x)(x-r)+P(r)$ ,

Where $Q(x)$ is a polynomial with degree one less than $P(x)$ .

Calculation:

Consider ,

  $f(x)=0.00000056x^4-0.000018x^3-0.016x^2+1.38x-0.38$

By using Remainder Theorem, $f(50)$ should be the remainder when the polynomial is divided by $x-50$ .

  $\begin{array}{l}\overline{)50}\begin{array}{cccc}0.00000056& -0.000018& \begin{array}{cc}-0.016& \begin{array}{cc}1.38& -0.38\end{array}\end{array}& \end{array}\\ \begin{array}{cccc}0.000028& 0.0005& -\text{0}\text{.775}& \text{ 30}\text{.25}\end{array}\\ \overline{\begin{array}{cccc}0.00000056& 0.00001& -0.0155& \text{ 0}\text{.605}\end{array}\hspace{.2em}|}\text{ 29}\text{.87}\end{array}$

The remainder is $29.87$ .

Therefore, by synthetic substitution, $f(50)=\mathrm{29.87.}$

Hence, the fuel economy when the car is travelling 50 miles per hour is $29.87$ miles per gallon.

c.

To determine

To calculate: the fuel economy when the car is travelling 60 miles per hour.

Answer to Problem 16PPS

The fuel economy is $28.1896$ miles per gallon.

Explanation of Solution

Given information:

A specific car’s fuel economy in miles per gallon can be $f(x)=0.00000056x^4-0.000018x^3-0.016x^2+1.38x-0.38$ where x represents the car’s speed in miles per hour.

Formula used:

Synthetic Substitution :- Synthetic division is used to find the value of a function.

Remainder Theorem:- If a polynomial $P(x)$ is divided by $x-r$ , the remainder s a constant $P(r)$ , and

  $P(x)=Q(x)(x-r)+P(r)$ ,

Where $Q(x)$ is a polynomial with degree one less than $P(x)$ .

Calculation:

Consider ,

  $f(x)=0.00000056x^4-0.000018x^3-0.016x^2+1.38x-0.38$

By using Remainder Theorem, $f(60)$ should be the remainder when the polynomial is divided by $x-60$ .

  $\begin{array}{l}\overline{)60}\begin{array}{cccc}0.00000056& -0.000018& \begin{array}{cc}-0.016& \begin{array}{cc}1.38& -0.38\end{array}\end{array}& \end{array}\\ \begin{array}{cccc}0.0000336& 0.000936& -\text{0}\text{.90384}& \text{ 28}\text{.5696}\end{array}\\ \overline{\begin{array}{cccc}0.00000056& 0.0000156& -0.015064& \text{ 0}\text{.47616}\end{array}\hspace{.2em}|}\text{ 28}\text{.1896}\end{array}$

The remainder is $28.1896$ .

Therefore, by synthetic substitution, $f(60)=\mathrm{28.1896.}$

Hence, the fuel economy when the car is travelling 60 miles per hour is $28.1896$ miles per gallon.