Chapter 1.4, Problem 85E

To determine

Find the matching and determining Constants.

Answer to Problem 85E

The value of $c$ for $g\left(x\right)is\boxed{-2}$

Explanation of Solution

Given information:

Match the data with one of the following functions and determine the value of the constant that will make the function fit the data in the table

  $f\left(x\right)=cx$ , $g\left(x\right)=c{x}^2$ , $h\left(x\right)=c\sqrt{\left|x\right|}$ and $r\left(x\right)=\frac{c}{x}$

  

Calculation:

Consider the given table.

  

Evaluate the given functions to satisfy the above data,

  $f\left(x\right)=cx$ , $g\left(x\right)=c{x}^2$ , $h\left(x\right)=c\sqrt{\left|x\right|}$ and $r\left(x\right)=\frac{c}{x}$

Consider $f\left(x\right)=cx$

  $\begin{array}{l}f\left(-4\right)=c\left(-4\right)\\ =-4c\end{array}$

This function can be equal to $-32$ if $c=8$

  $\begin{array}{l}f\left(1\right)=c\left(1\right)\\ =c\end{array}$

This function can be equal to $-2$ if $c=-2$

There is contradiction,

Hence the function cannot be matched with the data.

Now consider $g\left(x\right)=c{x}^2$

  $\begin{array}{l}g\left(-4\right)=c{\left(-4\right)}^2\\ =16c\end{array}$

This function can be equal to $-32$ if $c=-2$

  $\begin{array}{l}g\left(1\right)=c{\left(1\right)}^2\\ =c\end{array}$

This function can be equal to $-2$ if $c=-2$

Hence we get the required function.

The value of $c$ for $g\left(x\right)is\boxed{-2}$

Consider $h\left(x\right)=c\sqrt{\left|x\right|}$

  $\begin{array}{l}h\left(-4\right)=c\sqrt{\left|-4\right|}\\ =c\sqrt{4}\\ =2c\end{array}$

This function can be equal to $-32$ if $c=-16$

  $\begin{array}{l}h\left(1\right)=c\sqrt{\left|1\right|}\\ =c\end{array}$

This function can be equal to $-2$ if $c=-2$

This function cannot be matched with the data.

Now consider $r\left(x\right)=\frac{c}{x}$

  $r\left(-4\right)=\frac{c}{-4}$

This function can be equal to $-32$ if $c=128$

  $\begin{array}{l}r\left(1\right)=\frac{c}{1}\\ =c\end{array}$

This function can be equal to $-2$ if $c=-2$

This function cannot be matched with the data.

  $\begin{array}{l}f\left(2\right)={\left(2\right)}^2-\left(2\right)+1\\ =4-2+1\\ =4-1\\ =3\end{array}$

Now replace $xby\left(2+h\right)$

  $\begin{array}{l}f\left(2+h\right)={\left(2+h\right)}^2-\left(2+h\right)+1\\ =4+h^2+4h-1-h\\ =h^2+3h+3\end{array}$

Now substitute the calculated values in $\frac{f\left(2+h\right)-f\left(2\right)}{h},h\ne 0$

  $\begin{array}{l}\frac{f\left(2+h\right)-f\left(2\right)}{h}=\frac{h^2+3h+3-3}{h}\\ =\frac{h^2+3h}{h}=\frac{\cancel{h}\left(h+3\right)}{\cancel{h}}\\ \frac{f\left(2+h\right)-f\left(2\right)}{h}=h+3\end{array}$

Hence, the difference quotient is $\boxed{h+3}$