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Average Rate of Change - College Algebra

  1. A function f(x)f(x) passes through the points (−3,7)(-3, 7) and (5,−1)(5, -1). If g(x)=f(x−2)+3g(x) = f(x - 2) + 3, what is the average rate of change of g(x)g(x) over the interval [−1,7][-1, 7]?
    Answer: −1-1 First, find the average rate of change of f(x)f(x) over [−3,5][-3, 5]: f(5)−f(−3)5−(−3)=−1−78=−1\frac{f(5) - f(-3)}{5 - (-3)} = \frac{-1 - 7}{8} = -1. For g(x)=f(x−2)+3g(x) = f(x - 2) + 3, we need g(−1)g(-1) and g(7)g(7). Since g(−1)=f(−3)+3=7+3=10g(-1) = f(-3) + 3 = 7 + 3 = 10 and g(7)=f(5)+3=−1+3=2g(7) = f(5) + 3 = -1 + 3 = 2, the average rate of change is 2−107−(−1)=−88=−1\frac{2 - 10}{7 - (-1)} = \frac{-8}{8} = -1. Choice B results from incorrectly using the original interval length of 8 in the denominator twice. Choice C is the negative of the correct answer. Choice D ignores the negative sign from the calculation.
  2. The function k(x)k(x) has an average rate of change of −3-3 over the interval [2,6][2, 6] and an average rate of change of 55 over the interval [6,10][6, 10]. If k(2)=8k(2) = 8, what is k(8)k(8) assuming k(x)k(x) is linear on each interval?
    Answer: 66 Since k(x)k(x) has average rate of change −3-3 over [2,6][2,6] and k(2)=8k(2) = 8, we have k(6)−k(2)6−2=−3\frac{k(6) - k(2)}{6 - 2} = -3, so k(6)−84=−3\frac{k(6) - 8}{4} = -3, giving k(6)−8=−12k(6) - 8 = -12, thus k(6)=−4k(6) = -4. Since k(x)k(x) is linear on [6,10][6,10] with rate 55, we have k(8)=k(6)+5⋅(8−6)=−4+5⋅2=−4+10=6k(8) = k(6) + 5 \cdot (8-6) = -4 + 5 \cdot 2 = -4 + 10 = 6. Choice A would result from calculation errors. Choice B incorrectly uses the rate from the first interval. Choice D is k(6)+14k(6) + 14, a common arithmetic error.

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