Chapter 7.6, Problem 34PPE

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 34PPE

Rangeover given domain: $\left\{100,10,1,\frac{1}{10},\frac{1}{100},\frac{1}{1000}\right\}$

Range decreases as domain increases.

Explanation of Solution

Given information:

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

Exponential function: $h\left(x\right)={0.1}^x$

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 $h\left(x\right)={0.1}^x$ over the given domain, take x values from domain one at a time and evaluate y for one.

For x = −2

  $\begin{array}{c}h\left(x\right)={0.1}^{-2}\\ =\frac{1}{{0.1}^2}\\ =\frac{1}{0.1\times 0.1}\\ =\frac{1}{0.01}\\ =100\end{array}$

For x = −1

  $\begin{array}{c}h\left(x\right)={0.1}^{-1}\\ =\frac{1}{0.1}\\ =10\end{array}$

For x = 0

  $\begin{array}{c}h\left(x\right)={0.1}^0\\ =1\end{array}$

For x = 1

  $\begin{array}{c}h\left(x\right)={0.1}^1\\ =0.1\\ =\frac{1}{10}\end{array}$

For x = 2

  $\begin{array}{c}h\left(x\right)={0.1}^2\\ =0.1\times 0.1\\ =0.01\\ =\frac{1}{100}\end{array}$

For x = 3

  $\begin{array}{c}h\left(x\right)={0.1}^3\\ =0.1\times 0.1\times 0.1\\ =0.001\\ =\frac{1}{1000}\end{array}$

Thus Range is $\left\{100,10,1,\frac{1}{10},\frac{1}{100},\frac{1}{1000}\right\}$ .

Hence range decreases as domain increases.