Chapter 1.6, Problem 44E

(a)

To determine

To graph: The mathematical model function $\begin{array}{l}f\left(x\right)=\left\{\begin{array}{l}-1.97x+26.3\\ 0.505x^2-1.47x+6.3\text{'}\end{array}\right\}\\ \end{array}$ .

Explanation of Solution

Given information:

The given function is $\begin{array}{l}f\left(x\right)=\left\{\begin{array}{l}-1.97x+26.3\\ 0.505x^2-1.47x+6.3\text{'}\end{array}\right\}\\ \end{array}$

Graph:

Consider the function $\begin{array}{l}f\left(x\right)=\left\{\begin{array}{l}-1.97x+26.3\\ 0.505x^2-1.47x+6.3\text{'}\end{array}\right\}\\ \end{array}$

The graph for the given function $\begin{array}{l}f\left(x\right)=\left\{\begin{array}{l}-1.97x+26.3\\ 0.505x^2-1.47x+6.3\text{'}\end{array}\right\}\\ \end{array}$ is shown in figure-(1)

  

Figure-(1)

Therefore, the help of the above graph, the domain of $f\left(x\right)=0.505x^2-1.47x+6.3$ is on interval $\left(1,6\right)$ and the domain of $f\left(x\right)=-1.97x+26.3$ is on interval $\left(6.12\right)$ .

(b)

To determine

To find: The $f\left(5\right)$ and $f\left(11\right)$ .

Answer to Problem 44E

The function $f\left(5\right)=11.575$ and $f\left(11\right)=4.63$ .

Explanation of Solution

Given information:

The given function is $\begin{array}{l}f\left(x\right)=\left\{\begin{array}{l}-1.97x+26.3\\ 0.505x^2-1.47x+6.3\text{'}\end{array}\right\}\\ \end{array}$ .

Calculation:

Calculate the functions,

  $\begin{array}{c}f\left(x\right)=0.505x^2-1.47x+6.3\\ f\left(x\right)=0.505{\left(5\right)}^2-1.47\left(5\right)+6.3\\ =11.575\end{array}$

Therefore the function $f\left(5\right)=11.575$

  $\begin{array}{c}f\left(x\right)=-1.97x+26.3\\ f\left(x\right)=-1.97\left(11\right)+26.3\\ =4.63\end{array}$

Therefore the function $f\left(11\right)=4.63$ .