Chapter 7.6, Problem 39PPE

To determine

To evaluate: The given exponential function over a given domain and state if the values of range increase or decrease as domain increases.

Answer to Problem 39PPE

Range over given domain: $\left\{\frac{100}{0.09},\frac{100}{0.3},100,30,9,2.7\right\}$

Range decreases as domain increases.

Explanation of Solution

Given information:

Domain: $\left\{-2,-1,0,1,2,3\right\}$

Exponential function: $y=100\cdot {0.3}^x$

Concept Used:

For a given function $f\left(x\right)$ represented by y , domain is all possible values that x can take and range is all possible values of $f\left(x\right)$ or y .

Calculation:

To evaluate the function $y=100\cdot {0.3}^x$ over the given domain, take x values from domain one at a time and evaluate y for one.

For x = −2

  $\begin{array}{c}y=100\cdot {0.3}^{-2}\\ =100\cdot \frac{1}{0.3\times 0.3}\\ =100\cdot \frac{1}{0.09}\\ =\frac{100}{0.09}\end{array}$

For x = −1

  $\begin{array}{c}y=100\cdot {0.3}^{-1}\\ =\frac{100}{0.3}\end{array}$

For x = 0

  $\begin{array}{c}y=100\cdot {0.3}^0\\ =100\end{array}$

For x = 1

  $\begin{array}{c}y=100\cdot {0.3}^1\\ =30\end{array}$

For x = 2

  $\begin{array}{c}y=100\cdot {0.3}^2\\ =100\times 0.09\\ =9\end{array}$

For x = 3

  $\begin{array}{c}g\left(x\right)=100\cdot {0.3}^3\\ =100\times 0.027\\ =2.7\end{array}$

Thus Range is $\left\{\frac{100}{0.09},\frac{100}{0.3},100,30,9,2.7\right\}$ .

Hence range decreases as domain increases.