We can find (fg)(-2) using the fact that (fg)(-2) = f(-2)· g(-2). f(x) = x² + x f(-2) = (-2)² + (-2) = 4 – 2 = 2 g(x) = x – 5 g(-2) = -2 – 5 = -7 Thus, (fg)(-2) = f(-2) •g(-2) = 2(-7) = -14. We could also have found (fg)(-2) by multiplying f(x) · g(x) and then substituting -2 into the product. We will discuss how to multiply expressions such as x2 + x and x - 5 in Chapter 5. I

College Algebra
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ISBN:9781337282291
Author:Ron Larson
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ChapterP: Prerequisites
SectionP.4: Factoring Polynomials
Problem 84E: The rate of change of an autocatalytic chemical reaction is kQxkx2 where Q is the amount of the...
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We can find (fg)(-2) using the fact that
(fg)(-2) = f(-2)· g(-2).
f(x) = x² + x
f(-2) = (-2)² + (-2) = 4 – 2 = 2
g(x) = x – 5
g(-2) = -2 – 5 = -7
Thus,
(fg)(-2) = f(-2) •g(-2) = 2(-7) = -14.
We could also have found (fg)(-2) by multiplying f(x) · g(x) and then substituting
-2 into the product. We will discuss how to multiply expressions such as x2 + x and
x - 5 in Chapter 5. I
Transcribed Image Text:We can find (fg)(-2) using the fact that (fg)(-2) = f(-2)· g(-2). f(x) = x² + x f(-2) = (-2)² + (-2) = 4 – 2 = 2 g(x) = x – 5 g(-2) = -2 – 5 = -7 Thus, (fg)(-2) = f(-2) •g(-2) = 2(-7) = -14. We could also have found (fg)(-2) by multiplying f(x) · g(x) and then substituting -2 into the product. We will discuss how to multiply expressions such as x2 + x and x - 5 in Chapter 5. I
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