Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
6. Formalizing Relations and Functions
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Exercise 28 Page 272

The machine cannot run partial cycles.

Domain: { x | x is an integer, 0 ≤ x ≤ 23}
Range: See solution.
Is the Function Linear? Yes
Graph:

Practice makes perfect

We are given some information about a certain machine.

  • Warm-up time: 10 min
  • Running one cycle time: 15 min
  • Operation time including warm-up: 6 hours

We are asked to draw a graph showing the total time the machine operates during a day as a function of the number of cycles it runs.

Domain and Range

Let t(x) be the total time in minutes, the machine operates to run x cycles of 15 minutes each. Recall that there is a warm-up time of 10 minutes. t(x)= 15x+ 10Note that x cannot be negative, since it is impossible to run a negative number of cycles. Furthermore, since the machine cannot run partial cycles, x is an integer. rcl & x & ↙ & & ↘ x≥ 0 & & Integer Now, we have that x≥ 0. Let's find the corresponding output of x=0 by substituting 0 for x in the above function rule.

t(x)=15x+10
t( 0)=15( 0)+10
â–¼
Evaluate right-hand side
t( 0)=0+10
t(0)=10

We found that t(0)=10. This means that the machine will only run for 10 minutes if no cycles are run. Moreover, we know that the machine can operate for as long as 6hours a day, which is equivalent to 6* 60= 360 minutes. By substituting 360 for t(x), we can determine the maximum number of cycles that the machine can run.

t(x)=15x+10
360=15x+10
â–¼
Solve for x
350=15x
15x=350
x = 350/15
x= 70/3
x = 23.333333 ...

Therefore, the maximum number of cycles the machine can run in 6 hours is 23, since partial cycles are not allowed. We can write the domain using the obtained values. Domain:& { x | x is an integer, 0 ≤ x ≤ 23} Let's construct a table to find the range values. Keep the domain in mind! Since the function is discrete, we must find the corresponding range for each value in our domain.

x 15x+10 t(x)=15x+10
0 15( 0)+10 10
1 15( 1)+10 25
2 15( 2)+10 40
3 15( 3)+10 55
4 15( 4)+10 70
5 15( 5)+10 85
6 15( 6)+10 100
7 15( 7)+10 115
8 15( 8)+10 130
9 15( 9)+10 145
10 15( 10)+10 160
11 15( 11)+10 175
12 15( 12)+10 190
13 15( 13)+10 205
14 15( 14)+10 220
15 15( 15)+10 235
16 15( 16)+10 250
17 15( 17)+10 265
18 15( 18)+10 280
19 15( 19)+10 295
20 15( 20)+10 310
21 15( 21)+10 325
22 15( 22)+10 340
23 15( 23)+10 355

Using the table, we can write the range. Range: { &10,25,40,55,70,85,100, &115,130,145,160,175, &190,205,220,235,250,265, &280,295,310,325, 340,355 }

Is the Function a Linear Function?

Let's now plot the obtained points. Again, keep in mind domain and range!

Since the points appear to be on a straight line, the function is a linear function.