Chapter 1.5, Problem 50E

Solution

a.

To Determine:

The inverse function of $y={\left(\frac{3^{1.7}}{30}\left(x-65\right)\right)}^{\frac{1}{1.7}}+1$ having the domain $\left[65,100\right]$ algebraically.

To State:

What can the inverse function be used for.

The inverse function is $f^{-1}\left(y\right)=30{\left(\frac{y-1}{3}\right)}^{1.7}+65$ .

The inverse function can be used to find the corresponding percentage score if the GPA $y$ is given.

Given:

The function $y={\left(\frac{3^{1.7}}{30}\left(x-65\right)\right)}^{\frac{1}{1.7}}+1$ having the domain $\left[65,100\right]$ is used to calculate the four point scale GPA from percentage score.

GPA is considered zero if percentage score is below $65$ .

Concepts Used:

The inverse of $y=f\left(x\right)$ can be obtained by isolating $x$ from the equation $y=f\left(x\right)$ such that it is expressed in terms of $y$ alone.

Isolating a variable from an equation.

Calculations:

Isolate the variable $x$ from the equation such that it is expressed in terms of $y$ only:

  

The equation $30{\left(\frac{y-1}{3}\right)}^{1.7}+65=x$ shows $x$ in terms of $y$ only. If the original equation is rewritten after naming $y=f\left(x\right)$ then the inverse is $f^{-1}\left(y\right)=30{\left(\frac{y-1}{3}\right)}^{1.7}+65$ .

Conclusion:

The inverse function is $f^{-1}\left(y\right)=30{\left(\frac{y-1}{3}\right)}^{1.7}+65$ .

The inverse function can be used to find the corresponding percentage score if the GPA $y$ is given.

b.

To Determine:

If the domain of the inverse function has any restrictions.

Given:

The function $y={\left(\frac{3^{1.7}}{30}\left(x-65\right)\right)}^{\frac{1}{1.7}}+1$ having the domain $\left[65,100\right]$ is used to calculate the four point scale GPA from percentage score.

GPA is considered zero if percentage score is below $65$ .

Known from previous part:

The inverse function is $f^{-1}\left(y\right)=30{\left(\frac{y-1}{3}\right)}^{1.7}+65$ .

Concepts Used:

Non integral exponents are not defined over negative bases.

The domain of a function limits the range of its inverse.

Calculations:

The inverse function $f^{-1}\left(y\right)=30{\left(\frac{y-1}{3}\right)}^{1.7}+65$ involves the base $\frac{y-1}{3}$ raised to non integer exponent $1.7$ . Such a base cannot be negative. Find the equivalent condition on the input $y$ to the inverse function.

  $\begin{array}{l}\frac{y-1}{3}\ge 0\\ y-1\ge 0\\ y\ge 1\end{array}$

Thus, the input $y$ to the inverse function must be at least $1$ .

Also since the domain of the original function $f\left(x\right)={\left(\frac{3^{1.7}}{30}\left(x-65\right)\right)}^{\frac{1}{1.7}}+1$ is given to be $\left[65,100\right]$ , the range of the inverse is limited to $\left[65,100\right]$ . So no such inputs $y$ must be allowed in the domain of $f^{-1}$ which make the result not lie in $\left[65,100\right]$ . Hence the upper limit to $y$ is $f\left(100\right)={\left(\frac{3^{1.7}}{30}\left(100-65\right)\right)}^{\frac{1}{1.7}}+1$ . This reduces approximately to $4.28$ .

Conclusion:

The domain of the inverse function $f^{-1}\left(y\right)=30{\left(\frac{y-1}{3}\right)}^{1.7}+65$ is $\left[1,4.28\right]$ . Here the upper limit is an approximate value of $f\left(100\right)$ .

c.

To Verify Graphically:

The function found in part a. is indeed the inverse of the given function $f\left(x\right)={\left(\frac{3^{1.7}}{30}\left(x-65\right)\right)}^{\frac{1}{1.7}}+1$ .

Given:

The function $y={\left(\frac{3^{1.7}}{30}\left(x-65\right)\right)}^{\frac{1}{1.7}}+1$ having the domain $\left[65,100\right]$ is used to calculate the four point scale GPA from percentage score.

GPA is considered zero if percentage score is below $65$ .

Known from previous part:

The inverse function is $f^{-1}\left(y\right)=30{\left(\frac{y-1}{3}\right)}^{1.7}+65$ .

Concepts Used:

Graphing a function using a graphing calculator.

The graph of the inverse of a function is the original graph reflected about the line $y=x$ .

Graphs:

  

The graph of the inverse function obtained algebraically is shown in red in the above figure.

  

The graph of inverse obtained by reflecting the original function’s graph about the line $y=x$ is shown in blue in the above figure.

Conclusion:

The graphs of the inverse function obtained using the two methods are found to be identical. Thus the algebraically found inverse function $f^{-1}\left(y\right)=30{\left(\frac{y-1}{3}\right)}^{1.7}+65$ is verified graphically to be the inverse of the given function $f\left(x\right)={\left(\frac{3^{1.7}}{30}\left(x-65\right)\right)}^{\frac{1}{1.7}}+1$ .