Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
6. Function Operations
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Exercise 45 Page 402

Practice makes perfect
a We know that a computer store offers a 5 % discount off of the list price x for any computer bought with cash. Let f(x) model the price after the 5 % cash discount.

f(x)=x-0.05x ⇔ f(x)=0.95x Given the list price x, we can find the discounted price using the function above.

b We also know that the manufacturer offers a $ 200 rebate for each purchase of a computer. Therefore, the function g(x) that gives the price after this rebate is written as follows.

g(x)=x-200

c Assuming the list price of a computer is $1500, we want to find the price of the computer in which the discount is applied before the rebate. To find it, we need to use f(1500) as the input for g(x). Now, let's evaluate f(1500).
f(x)=0.95x
f( 1500)=0.95( 1500)
f(1500)=1425
Next, we will use f(1500)=1425 as the input for g(x).
g(x)=x-200
g( 1425)= 1425-200
g(1425)=1225
Therefore, g(f(1500))=1225, the price of the computer in which the discount is applied before the rebate.
d Conversely, we will now use g(1500) as the input for f(x). Now, let's evaluate g(1500).
g(x)=x-200
g( 1500)= 1500-200
g(1500)=1300
Next, we will use g(1500)=1300 as the input for f(x).
f(x)=0.95x
f( 1300)=0.95( 1300)
f(1300)=1235
Therefore, f(g(1500))=1235, the price of the computer in which the rebate is applied before the discount.