Houghton Mifflin Harcourt Algebra 1, 2015
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Houghton Mifflin Harcourt Algebra 1, 2015 View details
5. Comparing Properties of Linear Functions
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Exercise 15 Page 236

Practice makes perfect
a Let's find the function which represents Gillian's pay for t hours first. Then we can compare the domain and range of both functions.

Gillian's pay

Since Gillian earns $12.50 per hour, for t hours she will earn 12.50t dollars. Therefore, we can write the function which represents her pay for t hours as: g(t)=12.50t.

Domain

Domain is the possible inputs of a function. We know that Gillian works from 20 to 30 hours per week. Therefore, the domain of g(t) is the set of real numbers t, where 20 ≤ t ≤ 30. We also know that Emily's pay is given by the function e(t)=13t, where 15 ≤ t ≤ 25. This is the domain of e(t) because these are all the possible values of t.

Range

Since both g(t) and e(t) are linear functions with the domains 20 ≤ t ≤ 30 and 15 ≤ t ≤ 25, respectively, their ranges will be all real numbers from g(20)≤ g(t)≤ g(30) and e(15)≤ e(t)≤ e(25). Let's find g(20), g(30), e(15), and e(25) by substituting the given values of t into the functions!

g(20) and g(30)

t g(t)=12.50t g(t)
20 g( 20)=12.50 * 20 250
30 g( 30)=12.50 * 30 375

e(15) and e(25)

t e(t)=13t e(t)
15 e( 15)=13 * 15 195
25 e( 25)=13 * 25 325

Thus, the range of g(t) is the set of all real numbers g(t), where 250 ≤ g(t) ≤ 375 and the range of e(t) is the set of all real numbers e(t), where 195 ≤ e(t) ≤ 325.

b Let's compare their hourly wages first, and then we can compare the amount they earn per week.

Hourly wages

We know that Gillian earns $12.50 per hour. From the function e(t)=13t, we know that Emily earns $13 per hour. Therefore, Emily's hourly wage is greater than Gillian's and, hence, Gillian earns less per hour than Emily.

Amount earned per week

From Part A, we know that the ranges of g(t) and e(t) are 250 ≤ g(t) ≤ 375 and 195 ≤ e(t) ≤ 325, respectively. Therefore, Gillian earns between $250 and $375 per week while Emily earns between $195 and $325 per week.