AP Precalculus Quiz: Rates Of Change
20 questions · exam conditions
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Rates Of ChangeQuestion 1 of 20

The population of a species of fish in a lake is modeled by the polynomial function P(t)P(t), where tt is the time in years since the beginning of 2010. The average rate of change of P(t)P(t) over the interval [2,5][2, 5] is −150-150.

Based on the model, which of the following is the correct interpretation of the average rate of change?

From the beginning of 2012 to the beginning of 2015, the fish population decreased on average by 150 fish per year.
From the beginning of 2012 to the beginning of 2015, the total fish population decreased by 150 fish.
At the beginning of 2015, the fish population was 150 less than it was at the beginning of 2012.
From the beginning of 2012 to the beginning of 2015, the fish population increased on average by 150 fish per year.
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AP Precalculus Quiz

AP Precalculus Quiz: Rates Of Change

Practice Rates Of Change in AP Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Rates Of Change, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The population of a species of fish in a lake is modeled by the polynomial function P(t)P(t), where tt is the time in years since the beginning of 2010. The average rate of change of P(t)P(t) over the interval [2,5][2, 5] is −150-150.

Based on the model, which of the following is the correct interpretation of the average rate of change?

  1. From the beginning of 2012 to the beginning of 2015, the fish population decreased on average by 150 fish per year. (correct answer)
  2. From the beginning of 2012 to the beginning of 2015, the total fish population decreased by 150 fish.
  3. At the beginning of 2015, the fish population was 150 less than it was at the beginning of 2012.
  4. From the beginning of 2012 to the beginning of 2015, the fish population increased on average by 150 fish per year.

Explanation: The average rate of change describes how one quantity changes, on average, with respect to another. The interval [2,5][2, 5] corresponds to the time from the start of 2012 to the start of 2015. A negative rate of −150-150 indicates an average decrease of 150 fish per year over that period.

Question 2

What is the average rate of change of the polynomial function f(x)=x3−2x+1f(x) = x^3 - 2x + 1 on the interval [−1,3][-1, 3]?

  1. 5 (correct answer)
  2. 10
  3. 20
  4. 4.5

Explanation: The average rate of change is calculated using the formula f(b)−f(a)b−a\frac{f(b) - f(a)}{b - a}. First, evaluate the function at the endpoints of the interval: f(3)=33−2(3)+1=27−6+1=22f(3) = 3^3 - 2(3) + 1 = 27 - 6 + 1 = 22 and f(−1)=(−1)3−2(−1)+1=−1+2+1=2f(-1) = (-1)^3 - 2(-1) + 1 = -1 + 2 + 1 = 2. Then, apply the formula: 22−23−(−1)=204=5\frac{22 - 2}{3 - (-1)} = \frac{20}{4} = 5.

Question 3

Referring to the context, rate of change means output change per 1-unit input. In a physics test track, a car's speed is modeled by the polynomial p(t)=−2t2+12tp(t)= -2t^2+12t (m/s) for 0≤t≤60\le t\le 6. The speed rises from t=0t=0 to t=3t=3 and then falls from t=3t=3 to t=6t=6, showing acceleration then deceleration. Over t=0t=0 to t=3t=3, the average rate of change is p(3)−p(0)3\frac{p(3)-p(0)}{3}. Over t=3t=3 to t=6t=6, the average rate of change is p(6)−p(3)3\frac{p(6)-p(3)}{3}. A sensor delay is modeled by the rational function r(t)=10t−3r(t)=\frac{10}{t-3}, which has a vertical asymptote at t=3t=3. As tt approaches 3, the delay grows without bound in magnitude. Far from t=3t=3, the delay approaches 0 seconds. Engineers compare these rates to decide when measurements are least reliable. Based on the passage, at which interval is the car's average rate of change in speed negative?

  1. From t=0t=0 to t=3t=3
  2. From t=3t=3 to t=6t=6 (correct answer)
  3. From t=0t=0 to t=6t=6
  4. From t=2t=2 to t=3t=3

Explanation: This question tests AP Precalculus understanding of rates of change in polynomial and rational functions. Rates of change measure how a function's output changes relative to its input, crucial for understanding function behavior over intervals. In the given scenario, the car's speed is modeled by p(t) = -2t² + 12t, which increases from t=0 to t=3 (reaching maximum at t=3) and decreases from t=3 to t=6. Choice B is correct because the average rate of change from t=3 to t=6 is [p(6)-p(3)]/3 = [0-18]/3 = -6, which is negative, indicating the car is decelerating. Choice A is incorrect because from t=0 to t=3, the rate is [p(3)-p(0)]/3 = [18-0]/3 = 6, which is positive. Encourage students to calculate average rates systematically using the formula (f(b)-f(a))/(b-a). Watch for: confusing increasing/decreasing function values with positive/negative rates of change.

Question 4

Referring to the context, rate of change is the average change in speed per second. A car's speed is modeled by the polynomial p(t)=−t2+10tp(t)=-t^2+10t (m/s) for 0≤t≤100\le t\le 10. The speed increases from t=0t=0 because the model starts at p(0)=0p(0)=0 and rises quickly. It later decreases because the negative quadratic term eventually outweighs the linear term. For example, p(3)=21p(3)=21 and p(7)=21p(7)=21, indicating the peak occurs between them. A second measurement is modeled by the rational function r(t)=50t−12+10r(t)=\frac{50}{t-12}+10, approaching 1010 as tt increases. Near t=12t=12, r(t)r(t) has a vertical asymptote and is not usable. Based on the passage, what does a negative average rate of change of p(t)p(t) on an interval indicate?

  1. The speed is decreasing over that time interval. (correct answer)
  2. The speed equals zero somewhere on that interval.
  3. The speed approaches a horizontal asymptote on that interval.
  4. The speed becomes undefined at a vertical asymptote on that interval.

Explanation: This question tests AP Precalculus understanding of rates of change in polynomial and rational functions. Rates of change measure how a function's output changes relative to its input, crucial for understanding function behavior over intervals. In the given scenario, students must interpret what a negative average rate of change means in the context of speed modeled by p(t) = -t² + 10t. Choice A is correct because a negative average rate of change means the function's output (speed) is decreasing over that time interval - the ending value is less than the starting value. Choice B is incorrect because a negative rate of change doesn't necessarily mean the function equals zero; it only indicates a decrease. Encourage students to connect mathematical concepts to physical interpretations: negative rate of change in speed means deceleration. Practice distinguishing between the sign of a function and the sign of its rate of change.

Question 5

Based on the passage, rate of change is the average change in output per unit input over an interval. In an engineering stress test, a deflection estimate is modeled by the polynomial p(x)=x3−3x2p(x)=x^3-3x^2 for 0≤x≤40\le x\le 4, where xx is load units. The model decreases from x=0x=0 to x=2x=2 and increases from x=2x=2 to x=4x=4, showing a turning point. A sensor correction uses the rational function r(x)=8x−2r(x)=\frac{8}{x-2}, which has a vertical asymptote at x=2x=2. Near x=2x=2, the correction becomes extremely large in magnitude and destabilizes readings. Far from x=2x=2, the correction approaches 0 and becomes small. Technicians compare behavior near the turning point and near the asymptote. Referring to the context, how does the polynomial's rate-related behavior at x=2x=2 differ from the rational's behavior there?

  1. Both become unbounded at x=2x=2, so neither has a usable rate
  2. The polynomial stays finite and changes direction, while the rational becomes unbounded (correct answer)
  3. The rational stays finite and changes direction, while the polynomial becomes unbounded
  4. Both approach 0 at x=2x=2, so both rates must be zero there

Explanation: This question tests AP Precalculus understanding of rates of change in polynomial and rational functions. Rates of change measure how a function's output changes relative to its input, crucial for understanding function behavior over intervals. In the given scenario, the polynomial p(x) = x³ - 3x² has a turning point at x=2 (where p'(x) = 3x² - 6x = 0), while the rational function r(x) = 8/(x-2) has a vertical asymptote there. Choice B is correct because at x=2, the polynomial p(2) = 8 - 12 = -4 remains finite and simply changes from decreasing to increasing, while r(x) becomes unbounded as x approaches 2. Choice C incorrectly reverses which function becomes unbounded. Encourage students to distinguish between turning points (finite) and asymptotes (unbounded). Practice analyzing critical points versus discontinuities. Watch for: confusing polynomial extrema with rational asymptotes.

Question 6

The polynomial function p(x)p(x) has a single local maximum at x=cx=c. Which statement best describes the function's rate of change immediately around this point?

  1. The rate of change is positive for x<cx < c and negative for x>cx > c. (correct answer)
  2. The rate of change is negative for x<cx < c and positive for x>cx > c.
  3. The rate of change is positive for both x<cx < c and x>cx > c.
  4. The rate of change is negative for both x<cx < c and x>cx > c.

Explanation: A local maximum occurs where a function stops increasing and starts decreasing. An increasing function has a positive rate of change, and a decreasing function has a negative rate of change. Therefore, immediately to the left of the maximum (x<cx < c), the rate of change is positive, and immediately to the right (x>cx > c), the rate of change is negative.

Question 7

A polynomial function p(x)p(x) has a point of inflection at x=ax=a. Which of the following must be true about the rate of change of p(x)p(x) at or near x=ax=a?

  1. The rate of change of p(x)p(x) is zero at x=ax=a.
  2. The rate of change of p(x)p(x) has a local extremum at x=ax=a. (correct answer)
  3. The rate of change of p(x)p(x) is positive on one side of x=ax=a and negative on the other.
  4. The rate of change of p(x)p(x) is undefined at x=ax=a.

Explanation: A point of inflection is where the concavity of the function changes. This corresponds to the rate of change switching from increasing to decreasing, or vice versa. When a function (in this case, the rate of change) switches from increasing to decreasing or vice versa, it has a local extremum (maximum or minimum).

Question 8

What is the average rate of change of the rational function r(x)=x+1x−2r(x) = \frac{x+1}{x-2} on the interval [3,5][3, 5]?

  1. −2-2
  2. 11
  3. −1-1 (correct answer)
  4. −25-\frac{2}{5}

Explanation: First, evaluate the function at the endpoints: r(5)=5+15−2=63=2r(5) = \frac{5+1}{5-2} = \frac{6}{3} = 2 and r(3)=3+13−2=41=4r(3) = \frac{3+1}{3-2} = \frac{4}{1} = 4. Then, calculate the average rate of change: r(5)−r(3)5−3=2−42=−22=−1\frac{r(5) - r(3)}{5 - 3} = \frac{2 - 4}{2} = \frac{-2}{2} = -1.

Question 9

Let f(x)=x3f(x) = x^3 and g(x)=10x2g(x) = 10x^2. Which statement accurately compares the average rates of change of f(x)f(x) and g(x)g(x) on the interval [20,21][20, 21]?

  1. The average rate of change of f(x)f(x) is greater than the average rate of change of g(x)g(x). (correct answer)
  2. The average rate of change of g(x)g(x) is greater than the average rate of change of f(x)f(x).
  3. The average rates of change of f(x)f(x) and g(x)g(x) are equal on this interval.
  4. The comparison cannot be made without knowing more specific function values.

Explanation: For f(x)f(x), the average rate of change is 213−20321−20=9261−8000=1261\frac{21^3 - 20^3}{21-20} = 9261 - 8000 = 1261. For g(x)g(x), the average rate of change is 10(212)−10(202)21−20=10(441)−10(400)=4410−4000=410\frac{10(21^2) - 10(20^2)}{21-20} = 10(441) - 10(400) = 4410 - 4000 = 410. Since 1261>4101261 > 410, the average rate of change of f(x)f(x) is greater.

Question 10

For the function p(x)=x4−8x2p(x) = x^4 - 8x^2, the rate of change at x=2x=2 is zero. Which of the following does this value indicate?

  1. The function has a value of zero at x=2x=2.
  2. The function has a point of inflection at x=2x=2.
  3. The function has a local extremum, such as a minimum or maximum, at x=2x=2. (correct answer)
  4. The function's average rate of change is zero on any interval containing x=2x=2.

Explanation: A rate of change of zero at a point indicates that the function is momentarily not increasing or decreasing. For a smooth function like a polynomial, this occurs at a local maximum, a local minimum, or a stationary inflection point. These are all types of local extrema in a broader sense.

Question 11

Based on the passage, rate of change is the average change in a quantity per unit change of the input. In a physics sprint test, a cart's speed is modeled by the polynomial p(t)=−t2+8tp(t)= -t^2+8t (m/s) for 0≤t≤80\le t\le 8. The cart speeds up until mid-run and then slows due to friction. The average rate of change on [0,4][0,4] is p(4)−p(0)4\frac{p(4)-p(0)}{4}, and on [4,8][4,8] it is p(8)−p(4)4\frac{p(8)-p(4)}{4}. A timing glitch is modeled by the rational function r(t)=4t−4r(t)=\frac{4}{t-4} with a vertical asymptote at t=4t=4. Near 4 seconds, the glitch spikes and makes readings unreliable. Away from 4 seconds, the glitch approaches 0 and becomes minor. Referring to the context, which statement correctly compares the cart's average rates on [0,4][0,4] and [4,8][4,8]?

  1. They are both negative because the speed is always decreasing
  2. They are both positive because the speed is always increasing
  3. The first is positive, and the second is negative (correct answer)
  4. The first is zero, and the second is positive

Explanation: This question tests AP Precalculus understanding of rates of change in polynomial and rational functions. Rates of change measure how a function's output changes relative to its input, crucial for understanding function behavior over intervals. In the given scenario, the polynomial p(t) = -t² + 8t models cart speed, which has a maximum at t=4 (found by setting p'(t) = -2t + 8 = 0). Choice C is correct because on [0,4], the average rate is [p(4)-p(0)]/4 = [16-0]/4 = 4 (positive, speeding up), while on [4,8], the rate is [p(8)-p(4)]/4 = [0-16]/4 = -4 (negative, slowing down). Choice A is incorrect because the speed increases then decreases, not always decreasing. Encourage students to calculate average rates systematically and connect them to physical motion. Practice identifying turning points in quadratic functions. Watch for: assuming monotonic behavior without checking.

Question 12

Referring to the context, rate of change is the change in output divided by the change in input. In an economics simulation, daily profit (in thousands of dollars) from producing xx units is modeled by the polynomial p(x)=−x2+10x−9p(x)= -x^2+10x-9. Profit increases for low production and decreases after overproduction raises costs. The average rate of change from x=1x=1 to x=4x=4 is p(4)−p(1)3\frac{p(4)-p(1)}{3}. A separate efficiency penalty is modeled by the rational function r(x)=12x−5r(x)=\frac{12}{x-5}, which has a vertical asymptote at x=5x=5. As production approaches 5 units, the penalty becomes extremely large in magnitude. For very large xx, the penalty approaches 0, but it remains negative when x>5x>5. Managers compare these behaviors to decide safe production targets. Based on the passage, what does the asymptote of r(x)r(x) indicate about the rate-related behavior near x=5x=5?​

  1. The penalty becomes 0 at x=5x=5, so change stops there
  2. The penalty's magnitude grows without bound as xx approaches 5 (correct answer)
  3. The penalty reaches its maximum finite value exactly at x=5x=5
  4. The penalty equals the polynomial profit at x=5x=5 by definition

Explanation: This question tests AP Precalculus understanding of rates of change in polynomial and rational functions. Rates of change measure how a function's output changes relative to its input, crucial for understanding function behavior over intervals. In the given scenario, the rational function r(x) = 12/(x-5) has a vertical asymptote at x=5, meaning the denominator equals zero there. Choice B is correct because as x approaches 5, the denominator approaches 0, causing the penalty's magnitude to grow without bound - this represents an infinite rate of change near the asymptote. Choice A is incorrect because r(5) is undefined, not zero; the function cannot equal zero at its asymptote. Encourage students to analyze vertical asymptotes as points where rational functions exhibit extreme behavior. Practice recognizing that asymptotes indicate unbounded growth, not maximum values or zeros. Watch for: misunderstanding what happens at vertical asymptotes.

Question 13

A car's velocity is modeled by the polynomial p(t)=−t3+6t2p(t)= -t^3+6t^2 (m/s) for 0≤t≤60\le t\le 6. Rate of change means the average change in velocity per second over an interval. From t=0t=0 to t=2t=2, the car speeds up as p(t)p(t) rises. From t=2t=2 to t=4t=4, the increase slows because the added velocity each second shrinks. From t=4t=4 to t=6t=6, velocity decreases, indicating deceleration. Engineers also track sensor error with r(t)=12t−3+8r(t)=\frac{12}{t-3}+8, which has a vertical asymptote at t=3t=3. Near t=3t=3, small time changes cause large jumps in r(t)r(t). Far from t=3t=3, r(t)r(t) levels toward 8, so its rate of change diminishes. Based on the passage, what is the average rate of change of p(t)p(t) on [4,6][4,6]?​

  1. −8-8 m/s per s
  2. −12-12 m/s per s
  3. −16-16 m/s per s (correct answer)
  4. 1616 m/s per s

Explanation: This question tests AP Precalculus understanding of rates of change in polynomial and rational functions. Rates of change measure how a function's output changes relative to its input, crucial for understanding function behavior over intervals. To find the average rate of change on [4,6], we calculate [p(6) - p(4)]/(6-4) where p(t) = -t³+6t². At t=4: p(4) = -64+96 = 32; at t=6: p(6) = -216+216 = 0, giving average rate = (0-32)/2 = -16 m/s per s. Choice C is correct because the calculation yields a decrease of 32 m/s over 2 seconds, which equals -16 m/s per second. Choice D is incorrect because it gives the magnitude without the negative sign, missing that this represents deceleration. Encourage students to maintain negative signs when computing rates of decreasing functions. Practice calculating rates of change for intervals where functions decrease.

Question 14

Based on the passage, rate of change is the average change in speed per second over a chosen interval. A car's speed is modeled by the polynomial p(t)=−t2+6t+8p(t)=-t^2+6t+8 (m/s) for 0≤t≤60\le t\le 6. The model increases at first, since p(0)=8p(0)=8 and p(2)=16p(2)=16 show acceleration. It then decreases after the peak because the negative quadratic term reduces the speed. For example, p(4)=16p(4)=16 and p(6)=8p(6)=8 show deceleration later. A separate drag-limited model is r(t)=12t−7+6r(t)=\frac{12}{t-7}+6, approaching 66 as tt increases. The rational model is undefined at t=7t=7 due to a vertical asymptote. Referring to the context, how does the rate of change of r(t)r(t) differ near its vertical asymptote compared with large tt?​

  1. It changes very rapidly near t=7t=7 but changes slowly as tt becomes large. (correct answer)
  2. It changes slowly near t=7t=7 but changes rapidly as tt becomes large.
  3. It is constant near t=7t=7 and constant for large tt.
  4. It reaches a maximum at t=7t=7 because asymptotes are turning points.

Explanation: This question tests AP Precalculus understanding of rates of change in polynomial and rational functions. Rates of change measure how a function's output changes relative to its input, crucial for understanding function behavior over intervals. In the given scenario, the rational function r(t) = 12/(t-7) + 6 has a vertical asymptote at t=7, causing dramatic changes in function values near this point. Choice A is correct because near the vertical asymptote at t=7, small changes in t produce large changes in r(t), resulting in rapid rates of change, while as t becomes large, r(t) approaches 6 and changes slowly. Choice B incorrectly reverses this behavior - rational functions change most rapidly near their vertical asymptotes, not at large t values. Encourage students to visualize the graph of rational functions near vertical asymptotes versus their horizontal asymptotic behavior. Practice analyzing how proximity to asymptotes affects rates of change.

Question 15

Referring to the context, rate of change means average change in speed per second over a stated interval. A car's speed is modeled by the polynomial p(t)=−0.5t2+5t+4p(t)=-0.5t^2+5t+4 (m/s) for 0≤t≤100\le t\le 10. The speed increases early, since values like p(0)=4p(0)=4 and p(4)=16p(4)=16 show acceleration. Later, the speed decreases because the negative quadratic term reduces speed growth. A second sensor follows the rational model r(t)=18t−6+9r(t)=\frac{18}{t-6}+9, which approaches 99 as tt increases. The rational model has a vertical asymptote at t=6t=6, where the expression is undefined. Based on the passage, how does the rate of change of r(t)r(t) behave as tt gets large?

  1. It approaches 0 because the function levels off toward its horizontal asymptote. (correct answer)
  2. It increases without bound because the numerator stays constant.
  3. It becomes infinite because the function approaches the vertical asymptote.
  4. It stays constant because rational functions have constant average rates of change.

Explanation: This question tests AP Precalculus understanding of rates of change in polynomial and rational functions. Rates of change measure how a function's output changes relative to its input, crucial for understanding function behavior over intervals. In the given scenario, the rational function r(t) = 18/(t-6) + 9 approaches the horizontal asymptote y = 9 as t increases, which affects its rate of change behavior. Choice A is correct because as t gets large, r(t) approaches its horizontal asymptote of 9, meaning the function levels off and its rate of change approaches 0. Choice C is incorrect because it confuses the behavior near the vertical asymptote (at t=6) with the behavior as t approaches infinity. Encourage students to visualize how rational functions flatten out as they approach horizontal asymptotes. Practice analyzing long-term behavior of rational functions and connecting this to diminishing rates of change.

Question 16

Referring to the context, rate of change is measured by the slope between two points on a model. In a biology tank, nutrient level is approximated by the polynomial p(t)=−0.5t2+3t+2p(t)= -0.5t^2+3t+2 for 0≤t≤60\le t\le 6 (days). It increases early and then decreases after the peak as nutrients get consumed. The average rate of change from t=2t=2 to t=4t=4 is p(4)−p(2)2\frac{p(4)-p(2)}{2}. A probe error is modeled by the rational function r(t)=3t−3r(t)=\frac{3}{t-3} with a vertical asymptote at t=3t=3. Very close to day 3, the error becomes extremely large and can mask real trends. Far from day 3, the error approaches 0 and becomes negligible. Based on the passage, which interpretation of the asymptote is mathematically accurate?

  1. At t=3t=3, r(t)r(t) has a highest value that the error cannot exceed
  2. As tt nears 3, r(t)r(t) grows without bound in magnitude (correct answer)
  3. At t=3t=3, r(t)r(t) equals 0, so the error disappears instantly
  4. As tt nears 3, r(t)r(t) becomes constant, so the error stabilizes

Explanation: This question tests AP Precalculus understanding of rates of change in polynomial and rational functions. Rates of change measure how a function's output changes relative to its input, crucial for understanding function behavior over intervals. In the given scenario, the rational function r(t) = 3/(t-3) has a vertical asymptote at t=3, meaning the function becomes unbounded there. Choice B is correct because as t approaches 3, the denominator approaches 0, causing |r(t)| to grow without bound - this is the mathematical definition of a vertical asymptote. Choice C is incorrect because r(3) is undefined (division by zero), not equal to 0; the error becomes infinite, not zero. Encourage students to understand asymptotes as points of unbounded behavior, not zeros or maxima. Practice recognizing the mathematical meaning of vertical asymptotes. Watch for: confusing undefined values with zero values.

Question 17

The average rates of change of a polynomial function q(x)q(x) over consecutive equal-length input-value intervals can be described by a linear function with a non-zero slope. Which of the following must be the most specific classification for q(x)q(x)?

  1. q(x)q(x) is a linear function.
  2. q(x)q(x) is a quadratic function. (correct answer)
  3. q(x)q(x) is a cubic function.
  4. The degree of q(x)q(x) cannot be determined.

Explanation: For a polynomial, if the average rates of change over consecutive equal-length intervals form a linear pattern (i.e., the second differences of the function values are constant and non-zero), the function must be quadratic. A linear function would have constant average rates of change, and a cubic function's average rates of change would form a quadratic pattern.

Question 18

A polynomial function ff has a positive rate of change, and that rate of change is decreasing for all xx on the interval (a,b)(a, b). Which statement accurately describes the graph of ff on the interval (a,b)(a, b)?

  1. The graph of ff is increasing and concave up.
  2. The graph of ff is increasing and concave down. (correct answer)
  3. The graph of ff is decreasing and concave up.
  4. The graph of ff is decreasing and concave down.

Explanation: A positive rate of change means the function is increasing. A decreasing rate of change means the graph of the function is concave down. Therefore, the graph of ff is increasing and concave down on the interval.

Question 19

The volume of water being pumped into a tank is given by the function V(t)=2t2V(t) = 2t^2, where VV is the volume in gallons and tt is the time in minutes. The average rate of change of the volume on the interval [1,3][1,3] is calculated.

What is the correct interpretation of this average rate of change?

  1. Over the first three minutes, the volume increased by an average of 8 gallons.
  2. Between minute 1 and minute 3, the volume increased by an average of 8 gallons per minute. (correct answer)
  3. At minute 3, the water was flowing into the tank at a rate of 8 gallons per minute.
  4. The total increase in volume between minute 1 and minute 3 was 8 gallons.

Explanation: First, calculate the average rate of change: V(3)=2(32)=18V(3) = 2(3^2) = 18 and V(1)=2(12)=2V(1) = 2(1^2) = 2. The average rate of change is 18−23−1=162=8\frac{18 - 2}{3 - 1} = \frac{16}{2} = 8. The units are gallons per minute. This value represents the average rate at which the volume increased over the time interval from t=1t=1 to t=3t=3 minutes.

Question 20

The average rate of change of a function g(x)g(x) over the interval [a,b][a, b] is geometrically represented by the slope of the secant line connecting the points (a,g(a))(a, g(a)) and (b,g(b))(b, g(b)). Which of the following expressions represents this value?

  1. g(b)+g(a)2\frac{g(b) + g(a)}{2}
  2. g(b)−g(a)b−a\frac{g(b) - g(a)}{b - a} (correct answer)
  3. g(b)−g(a)g(b) - g(a)
  4. b−ag(b)−g(a)\frac{b - a}{g(b) - g(a)}

Explanation: The slope of a line between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by the formula y2−y1x2−x1\frac{y_2 - y_1}{x_2 - x_1}. For the points (a,g(a))(a, g(a)) and (b,g(b))(b, g(b)) on the graph of g(x)g(x), this translates directly to g(b)−g(a)b−a\frac{g(b) - g(a)}{b - a}, which is the definition of the average rate of change.