Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games


Sign up

Log in

Opening subject page...

Loading your content

Practice

  • All Subjects
  • Algebra Flashcards
  • SAT Math Practice Tests
  • Math Question of the Day
  • Live Classes
  • On-Demand Courses

Varsity Tutors

  • Find a Tutor
  • Test Prep
  • Online Classes
  • K-12 Learning
  • College Search
  • VarsityTutors.com

© 2026 Varsity Tutors. All rights reserved.

← Back to quizzes

AP Calculus AB Quiz

AP Calculus AB Quiz: Rate Of Change At A Point

Practice Rate Of Change At A Point in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

If f(x)f(x)f(x) is differentiable, which best describes the instantaneous rate of change at x=ax=ax=a?

Select an answer to continue

What this quiz covers

This quiz focuses on Rate Of Change At A Point, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

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

If f(x)f(x)f(x) is differentiable, which best describes the instantaneous rate of change at x=ax=ax=a?

  1. The slope of the secant line through (a,f(a))(a,f(a))(a,f(a)) and (a+1,f(a+1))(a+1,f(a+1))(a+1,f(a+1)).
  2. The change in fff from x=0x=0x=0 to x=ax=ax=a.
  3. The slope of the tangent line to y=f(x)y=f(x)y=f(x) at x=ax=ax=a. (correct answer)
  4. The average value of fff on [a,a+1][a,a+1][a,a+1].
  5. The value f(a)f(a)f(a).

Explanation: Instantaneous rate of change is the slope of the tangent at x=a, choice C, differing from average, which is the secant slope over an interval like in choice A. This tangent slope captures the rate precisely at a. Choice A is average over [a,a+1], and choice D is an average value, not rate. Students commonly confuse the function value (choice E) or total change (choice B) with the rate. The derivative embodies the instantaneous concept. For strategy, associate tangent slopes with instantaneous and secants with averages in graphical interpretations.

Question 2

For temperature T(t)T(t)T(t), which expression represents the instantaneous rate of change at t=at=at=a?

  1. T(a+1)−T(a)1\dfrac{T(a+1)-T(a)}{1}1T(a+1)−T(a)​
  2. T(b)−T(a)b−a\dfrac{T(b)-T(a)}{b-a}b−aT(b)−T(a)​
  3. lim⁡h→0T(a+h)−T(a)h\lim_{h\to 0}\dfrac{T(a+h)-T(a)}{h}limh→0​hT(a+h)−T(a)​ (correct answer)
  4. T(a+h)−T(a)T(a+h)-T(a)T(a+h)−T(a) for small hhh
  5. T(a)−T(0)a\dfrac{T(a)-T(0)}{a}aT(a)−T(0)​

Explanation: The instantaneous rate of change at t=a captures how rapidly the temperature is changing at that precise moment. Choice C correctly uses the limit definition, examining the rate as the time interval h approaches zero. Choice A approximates the instantaneous rate using a one-unit interval, but this gives the average rate over [a, a+1], not the exact instantaneous rate. Choice B explicitly calculates an average rate over the interval [a,b]. Choice D gives a temperature difference without dividing by time, so it's not a rate at all. Choice E calculates the average rate from time 0 to time a. The fundamental principle: instantaneous rates require limits to isolate the rate at a single point.

Question 3

If x(t)x(t)x(t) is a particle’s position, which expression gives instantaneous velocity at t=bt=bt=b?

  1. x(b)−x(a)b−a\dfrac{x(b)-x(a)}{b-a}b−ax(b)−x(a)​
  2. lim⁡h→0x(b+h)−x(b)h\lim_{h\to 0}\dfrac{x(b+h)-x(b)}{h}limh→0​hx(b+h)−x(b)​ (correct answer)
  3. x(b+h)−x(b)x(b+h)-x(b)x(b+h)−x(b)
  4. x(b+1)−x(b−1)2\dfrac{x(b+1)-x(b-1)}{2}2x(b+1)−x(b−1)​
  5. x(b)b\dfrac{x(b)}{b}bx(b)​

Explanation: Instantaneous velocity at t=bt=bt=b represents the particle's speed and direction at that exact moment. Choice B correctly uses the limit definition, where we examine the velocity as the time interval hhh approaches zero. Choice A calculates the average velocity over the interval [a,b][a, b][a,b], which tells us the overall motion between two times but not the specific velocity at t=bt=bt=b. Choice C gives a position difference without dividing by time, so it's not a velocity. Choice D approximates using a centered difference over a 2-unit interval, giving an average velocity. Choice E would only represent average velocity if the particle started at position 0 at time 0. Remember: instantaneous velocity requires a limit to isolate the rate at a single instant.

Question 4

Water volume is V(t)V(t)V(t). Which expression is the instantaneous inflow rate at t=10t=10t=10 minutes?​

  1. V(10)−V(0)10\displaystyle \frac{V(10)-V(0)}{10}10V(10)−V(0)​
  2. V(12)−V(10)2\displaystyle \frac{V(12)-V(10)}{2}2V(12)−V(10)​
  3. lim⁡h→0V(10+h)−V(10)h\displaystyle \lim_{h\to 0}\frac{V(10+h)-V(10)}{h}h→0lim​hV(10+h)−V(10)​ (correct answer)
  4. V(10+h)−V(10)\displaystyle V(10+h)-V(10)V(10+h)−V(10)
  5. V(15)−V(5)10\displaystyle \frac{V(15)-V(5)}{10}10V(15)−V(5)​

Explanation: The instantaneous inflow rate at a specific time is the derivative of volume with respect to time, found by taking the limit of average rates over vanishingly small intervals. Choice C correctly expresses this as the limit as h approaches 0 of [V(10+h)-V(10)]/h, which defines V'(10). Choice A calculates the average rate over the first 10 minutes, Choice B gives the average rate from t=10 to t=12, and Choice E gives the average rate from t=5 to t=15. Choice D is missing the crucial division by h, giving only a change in volume rather than a rate. The distinguishing feature of instantaneous rates is the limit process—without it, you only have average rates over finite intervals.

Question 5

A tank contains V(t)V(t)V(t) liters after ttt minutes. Which gives the instantaneous filling rate at t=10t=10t=10?​​

  1. V(10)−V(0)10\dfrac{V(10)-V(0)}{10}10V(10)−V(0)​
  2. lim⁡h→0V(10+h)−V(10)h\lim_{h\to 0}\dfrac{V(10+h)-V(10)}{h}limh→0​hV(10+h)−V(10)​ (correct answer)
  3. V(11)−V(10)V(11)-V(10)V(11)−V(10)
  4. V(12)−V(8)12−8\dfrac{V(12)-V(8)}{12-8}12−8V(12)−V(8)​
  5. V(15)−V(10)15−10\dfrac{V(15)-V(10)}{15-10}15−10V(15)−V(10)​

Explanation: The instantaneous filling rate represents how fast the volume is changing at exactly t=10t=10t=10 minutes, which requires the derivative at that point. Choice B correctly expresses this as lim⁡h→0V(10+h)−V(10)h\lim_{h\to 0}\frac{V(10+h)-V(10)}{h}limh→0​hV(10+h)−V(10)​, the limit definition of the derivative. Choices A, D, and E calculate average rates over various time intervals, telling us the average filling rate but not the instantaneous rate. Choice C computes a volume difference without dividing by time, giving units of liters rather than liters per minute. Students often confuse average rates (which use two distinct points) with instantaneous rates (which use a limit). The key distinction: instantaneous rates capture the rate at a single moment through limits, while average rates measure change over an interval.

Question 6

A tank’s volume is V(t)V(t)V(t) liters. Which quantity gives the instantaneous filling rate at t=10t=10t=10 minutes?​​​

  1. V(10)−V(0)10\displaystyle \frac{V(10)-V(0)}{10}10V(10)−V(0)​
  2. lim⁡h→0V(10+h)−V(10)h\displaystyle \lim_{h\to 0}\frac{V(10+h)-V(10)}{h}h→0lim​hV(10+h)−V(10)​ (correct answer)
  3. V(10)−V(9)\displaystyle V(10)-V(9)V(10)−V(9)
  4. V(12)−V(8)4\displaystyle \frac{V(12)-V(8)}{4}4V(12)−V(8)​
  5. The total amount of water added from t=0t=0t=0 to t=10t=10t=10, V(10)−V(0)V(10)-V(0)V(10)−V(0)

Explanation: To find the instantaneous filling rate at a specific time, we need the derivative at that point, not an average over any interval. Choice A gives the average rate over 10 minutes, choices C and D give rates over fixed intervals (1 minute and 4 minutes respectively), and choice E describes the total change, not a rate. Only choice B uses the limit definition of the derivative, which captures the instantaneous rate by considering what happens as the time interval approaches zero. Many students mistakenly think that V(10)-V(9) gives the instantaneous rate because it uses a small interval, but this is still an average rate over that 1-minute period. The key insight is that instantaneous rates require limits, while average rates use fixed intervals.

Question 7

For a function fff, which statement best characterizes the instantaneous rate of change at x=1x=1x=1?

  1. The slope of the secant line from x=1x=1x=1 to x=2x=2x=2
  2. The value f(1)f(1)f(1)
  3. The slope of the tangent line to y=f(x)y=f(x)y=f(x) at x=1x=1x=1 (correct answer)
  4. The average rate of change on [0,1][0,1][0,1]
  5. The total change f(2)−f(0)f(2)-f(0)f(2)−f(0)

Explanation: The instantaneous rate of change at x=1 describes how rapidly f(x) is changing at that exact point. Choice C correctly identifies this as the slope of the tangent line at x=1, which geometrically captures the instantaneous rate. Choice A describes the average rate of change over the interval [1,2], telling us the overall trend between those points but not the specific rate at x=1. Choice B gives the function value itself, not a rate of change. Choice D calculates the average rate from x=0 to x=1. Choice E gives the total change over a 2-unit interval, not a rate. The fundamental distinction: instantaneous rates use tangent lines (limits), while average rates use secant lines (difference quotients).

Question 8

If V(t)V(t)V(t) is water volume, which expression is the instantaneous filling rate at t=10t=10t=10 minutes?

  1. V(20)−V(0)20\dfrac{V(20)-V(0)}{20}20V(20)−V(0)​
  2. V(10)−V(9)V(10)-V(9)V(10)−V(9)
  3. V(12)−V(8)12−8\dfrac{V(12)-V(8)}{12-8}12−8V(12)−V(8)​
  4. lim⁡h→0V(10+h)−V(10)h\displaystyle \lim_{h\to 0}\dfrac{V(10+h)-V(10)}{h}h→0lim​hV(10+h)−V(10)​ (correct answer)
  5. V(10)−V(0)10\dfrac{V(10)-V(0)}{10}10V(10)−V(0)​

Explanation: The key distinction is that average rates average change over time periods, such as [V(b) - V(a)] / (b - a), while instantaneous rates capture the rate at an exact instant via the derivative limit. For V(t) at t=10, the instantaneous filling rate is the limit as h approaches 0 of [V(10+h) - V(10)] / h, matching option D. Options A, C, and E are average rates, and B is a volume difference without normalization by time. A frequent mix-up is thinking a tiny interval like B approximates it well enough, but it lacks the division and limit for true instantaneity. Always compare by looking for the limit form to identify instantaneous rates across contexts.

Question 9

For position s(t)s(t)s(t) of a runner, which expression represents the instantaneous velocity at t=3t=3t=3?

  1. s(5)−s(1)5−1\dfrac{s(5)-s(1)}{5-1}5−1s(5)−s(1)​
  2. s(3)−s(0)3−0\dfrac{s(3)-s(0)}{3-0}3−0s(3)−s(0)​
  3. lim⁡h→0s(3+h)−s(3)h\lim_{h\to 0}\dfrac{s(3+h)-s(3)}{h}limh→0​hs(3+h)−s(3)​ (correct answer)
  4. s(3)−s(2)s(3)-s(2)s(3)−s(2)
  5. s(4)−s(3)4−3\dfrac{s(4)-s(3)}{4-3}4−3s(4)−s(3)​

Explanation: The instantaneous velocity at a specific time is the rate of change of position at that exact moment, not over an interval. Choice C correctly uses the limit definition: as the time interval h approaches 0, we get the instantaneous rate at t=3. Choices A, B, and E calculate average velocities over intervals (like [1,5], [0,3], or [3,4]), which give the average speed during those time periods. Choice D simply finds a position difference without dividing by time, so it doesn't represent velocity at all. Remember: instantaneous rates always involve limits, while average rates use the difference quotient over a finite interval.

Question 10

For profit R(t)R(t)R(t), which quantity is the instantaneous rate of change of profit at time t=ct=ct=c?

  1. R(c+1)−R(c)R(c+1)-R(c)R(c+1)−R(c)
  2. R(c)−R(0)c\dfrac{R(c)-R(0)}{c}cR(c)−R(0)​
  3. R(c+2)−R(c−2)4\dfrac{R(c+2)-R(c-2)}{4}4R(c+2)−R(c−2)​
  4. lim⁡h→0R(c+h)−R(c)h\lim_{h\to 0}\dfrac{R(c+h)-R(c)}{h}limh→0​hR(c+h)−R(c)​ (correct answer)
  5. The total profit earned up to time ccc

Explanation: The instantaneous rate of change of profit at t=c represents how quickly profit is changing at that exact moment. Choice D correctly uses the limit definition, examining how the profit rate behaves as the time interval h shrinks to zero around time c. Choice A gives the profit change over one time unit, which is an average rate over [c, c+1], not the instantaneous rate. Choice B calculates the average profit rate from the beginning. Choice C finds an average rate over a 4-unit interval centered at c. Choice E describes total profit, not a rate of change. Remember: instantaneous rates capture the rate at a single point using limits, while finite differences give average rates.

Question 11

Let A(t)A(t)A(t) be area of an oil spill; which represents the instantaneous rate of change at t=7t=7t=7?

  1. A(9)−A(7)2\dfrac{A(9)-A(7)}{2}2A(9)−A(7)​
  2. A(7)−A(6)A(7)-A(6)A(7)−A(6)
  3. A(7)−A(0)7\dfrac{A(7)-A(0)}{7}7A(7)−A(0)​
  4. lim⁡h→0A(7+h)−A(7)h\lim_{h\to 0}\dfrac{A(7+h)-A(7)}{h}limh→0​hA(7+h)−A(7)​ (correct answer)
  5. The average value of AAA on [0,7][0,7][0,7]

Explanation: The instantaneous rate of change at t=7 tells us exactly how fast the oil spill area is expanding at that moment. Choice D correctly uses the limit definition, examining the rate as the time interval h approaches zero. Choice A calculates the average rate over the 2-hour interval [7,9], which gives the average expansion rate during that period, not the instantaneous rate at t=7. Choice B approximates using a one-hour interval but doesn't divide by time. Choice C gives the average rate from the beginning. Choice E refers to the average area value, not a rate of change. The key insight: instantaneous rates require limits to capture the exact rate at a single point in time, distinguishing them from average rates over intervals.

Question 12

The temperature is T(t)T(t)T(t). Which quantity represents the instantaneous rate of temperature change at t=2t=2t=2 hours?​

  1. T(4)−T(0)4\displaystyle \frac{T(4)-T(0)}{4}4T(4)−T(0)​
  2. lim⁡h→0T(2+h)−T(2)h\displaystyle \lim_{h\to 0}\frac{T(2+h)-T(2)}{h}h→0lim​hT(2+h)−T(2)​ (correct answer)
  3. T(2)−T(0)\displaystyle T(2)-T(0)T(2)−T(0)
  4. T(3)−T(1)2\displaystyle \frac{T(3)-T(1)}{2}2T(3)−T(1)​
  5. T(2+h)−T(2)2\displaystyle \frac{T(2+h)-T(2)}{2}2T(2+h)−T(2)​

Explanation: The instantaneous rate of temperature change at a specific time is the derivative of temperature with respect to time, found through the limit definition. Choice B correctly expresses this as the limit as h approaches 0 of [T(2+h)-T(2)]/h, which gives T'(2). Choice A calculates the average rate over the first 4 hours, Choice C is just a temperature difference without dividing by time, and Choice D gives the average rate from t=1 to t=3. Choice E incorrectly divides by 2 instead of h, failing to create a proper rate expression. The defining characteristic of instantaneous rates is the limit process that considers what happens as the time interval approaches zero.

Question 13

Let s(t)s(t)s(t) be a car’s position. Which expression represents the instantaneous velocity at t=3t=3t=3?​

  1. lim⁡h→0s(3+h)−s(3)h\displaystyle \lim_{h\to 0}\frac{s(3+h)-s(3)}{h}h→0lim​hs(3+h)−s(3)​ (correct answer)
  2. s(5)−s(3)5−3\displaystyle \frac{s(5)-s(3)}{5-3}5−3s(5)−s(3)​
  3. s(3)−s(0)\displaystyle s(3)-s(0)s(3)−s(0)
  4. s(3)−s(0)3\displaystyle \frac{s(3)-s(0)}{3}3s(3)−s(0)​
  5. s(3+h)−s(3)h\displaystyle \frac{s(3+h)-s(3)}{h}hs(3+h)−s(3)​ for a fixed nonzero hhh

Explanation: The instantaneous velocity at a specific time is the derivative of position, which is defined as the limit of the average velocity over smaller and smaller time intervals. Choice A correctly represents this as the limit as h approaches 0 of [s(3+h)-s(3)]/h, which is the definition of the derivative at t=3. Choice B gives the average velocity between t=3 and t=5, not the instantaneous velocity at t=3. Choices C and D involve the position at t=0, which is irrelevant for finding velocity at t=3. Choice E lacks the crucial limit, giving only an average velocity over a fixed interval h. Remember: instantaneous rates always require a limit as the interval approaches zero, while average rates use fixed intervals.

Question 14

For differentiable fff, which quantity gives the instantaneous rate of change of fff at x=2x=2x=2?​

  1. f(4)−f(0)4\displaystyle \frac{f(4)-f(0)}{4}4f(4)−f(0)​
  2. f(2)−f(1)\displaystyle f(2)-f(1)f(2)−f(1)
  3. lim⁡x→2f(x)−f(2)x−2\displaystyle \lim_{x\to 2}\frac{f(x)-f(2)}{x-2}x→2lim​x−2f(x)−f(2)​ (correct answer)
  4. f(3)−f(2)1\displaystyle \frac{f(3)-f(2)}{1}1f(3)−f(2)​
  5. f(2)−f(−2)4\displaystyle \frac{f(2)-f(-2)}{4}4f(2)−f(−2)​

Explanation: The instantaneous rate of change of a function at a point is precisely the derivative at that point, which is defined as a limit. Choice C correctly shows this as the limit as x approaches 2 of [f(x)-f(2)]/(x-2), which is the definition of f'(2). Choice A gives the average rate of change from x=0 to x=4, not the instantaneous rate at x=2. Choice B is simply a difference of function values without any division by the change in x. Choice D gives the average rate from x=2 to x=3, and Choice E gives the average rate from x=-2 to x=2. The key distinction is that instantaneous rates require taking a limit as the interval shrinks to zero, while average rates use fixed intervals.

Question 15

Let h(t)h(t)h(t) be height of a plant; which describes the instantaneous growth rate at t=4t=4t=4?

  1. The average growth rate on [0,4][0,4][0,4]
  2. The slope of the secant line through t=3t=3t=3 and t=5t=5t=5
  3. The slope of the tangent line to y=h(t)y=h(t)y=h(t) at t=4t=4t=4 (correct answer)
  4. The change in height h(5)−h(3)h(5)-h(3)h(5)−h(3)
  5. The average value of hhh on [3,5][3,5][3,5]

Explanation: The instantaneous growth rate at t=4 tells us exactly how fast the plant is growing at that specific moment. Choice C correctly identifies this as the slope of the tangent line to the height curve at t=4, which geometrically represents the instantaneous rate of change. Choice A describes the average growth rate over the first 4 time units. Choice B gives the slope of a secant line through points at t=3 and t=5, which represents the average growth rate over that 2-unit interval, not the instantaneous rate at t=4. Choice D is just a height difference, not a rate. Choice E refers to the average height value, not a growth rate. The key principle: tangent lines capture instantaneous behavior, secant lines capture average behavior.

Question 16

Let V(t)V(t)V(t) be water volume in a tank; which expression gives the instantaneous rate of change at t=10t=10t=10?

  1. lim⁡h→0V(10+h)−V(10)h\lim_{h\to 0}\dfrac{V(10+h)-V(10)}{h}limh→0​hV(10+h)−V(10)​ (correct answer)
  2. V(12)−V(8)12−8\dfrac{V(12)-V(8)}{12-8}12−8V(12)−V(8)​
  3. V(11)−V(10)V(11)-V(10)V(11)−V(10)
  4. V(10)−V(0)10−0\dfrac{V(10)-V(0)}{10-0}10−0V(10)−V(0)​
  5. V(10)V(10)V(10)

Explanation: The instantaneous rate of change represents how fast the water volume is changing at the exact moment t=10t=10t=10. Choice A correctly expresses this using the limit definition, where we examine the rate as the time interval hhh shrinks to 000. Choice B calculates the average rate over the 4-minute interval [8,12][8,12][8,12], telling us the average flow rate during that period, not the instantaneous rate at t=10t=10t=10. Choice C finds the change over one minute, which approximates but doesn't equal the instantaneous rate. Choice D gives the average rate from the beginning, and Choice E just gives the volume itself, not a rate. The key distinction: instantaneous rates require limits to capture the rate at a single point in time.

Question 17

Let g(x)g(x)g(x) be differentiable. Which expression represents the slope of the tangent line to y=g(x)y=g(x)y=g(x) at x=5x=5x=5?

  1. g(7)−g(3)7−3\dfrac{g(7)-g(3)}{7-3}7−3g(7)−g(3)​
  2. g(6)−g(5)6−5\dfrac{g(6)-g(5)}{6-5}6−5g(6)−g(5)​
  3. lim⁡h→0g(5+h)−g(5)h\displaystyle \lim_{h\to 0}\dfrac{g(5+h)-g(5)}{h}h→0lim​hg(5+h)−g(5)​ (correct answer)
  4. g(5+h)−g(5)g(5+h)-g(5)g(5+h)−g(5)
  5. g(5)−g(1)5−1\dfrac{g(5)-g(1)}{5-1}5−1g(5)−g(1)​

Explanation: The slope of a secant line gives the average rate over an interval, but the tangent slope at a point is the instantaneous rate, defined by the derivative limit. For g(x) at x=5, this is the limit as h approaches 0 of [g(5+h) - g(5)] / h, choice C. A, B, and E are secant slopes, and D is a difference without the quotient. A common error is equating small-interval averages like B to tangent slopes, but they only approximate without the limit. Compare expressions by checking for the limit process to reliably spot instantaneous rates in graphical or functional contexts.

Question 18

A savings account balance is B(t)B(t)B(t). Which expression represents the instantaneous earning rate at t=12t=12t=12 months?​​

  1. B(12)−B(0)12\dfrac{B(12)-B(0)}{12}12B(12)−B(0)​
  2. B(18)−B(6)18−6\dfrac{B(18)-B(6)}{18-6}18−6B(18)−B(6)​
  3. lim⁡h→0B(12+h)−B(12)h\lim_{h\to 0}\dfrac{B(12+h)-B(12)}{h}limh→0​hB(12+h)−B(12)​ (correct answer)
  4. B(12)−B(11)B(12)-B(11)B(12)−B(11)
  5. B(13)−B(12)13−12\dfrac{B(13)-B(12)}{13-12}13−12B(13)−B(12)​

Explanation: The instantaneous earning rate at t=12t=12t=12 months represents how fast the account balance is growing at that exact time, which is the derivative of the balance function. Choice C correctly expresses this as lim⁡h→0B(12+h)−B(12)h\lim_{h\to 0}\frac{B(12+h)-B(12)}{h}limh→0​hB(12+h)−B(12)​, capturing the precise rate of change through the limit definition of the derivative. Choices A and B calculate average earning rates over different time periods, while choices D and E approximate using one-month differences but without the limiting process. Students often confuse the earnings over one month (like B(12)−B(11)B(12)-B(11)B(12)−B(11)) with the instantaneous rate, but this gives an average rate for that month, not the rate at the instant t=12t=12t=12. The distinction: instantaneous rates use limits to find the rate at a single moment, while finite differences give average rates over intervals.

Question 19

Let C(x)C(x)C(x) be the total cost to produce xxx items. Which represents the instantaneous marginal cost at x=50x=50x=50?​​

  1. C(60)−C(40)60−40\dfrac{C(60)-C(40)}{60-40}60−40C(60)−C(40)​
  2. C(51)−C(50)C(51)-C(50)C(51)−C(50)
  3. lim⁡h→0C(50+h)−C(50)h\lim_{h\to 0}\dfrac{C(50+h)-C(50)}{h}limh→0​hC(50+h)−C(50)​ (correct answer)
  4. C(50)−C(0)50\dfrac{C(50)-C(0)}{50}50C(50)−C(0)​
  5. C(50)−C(49)C(50)-C(49)C(50)−C(49)

Explanation: The instantaneous marginal cost at x=50x=50x=50 represents the rate at which total cost changes when producing exactly 50 items, which is the derivative of the cost function. Choice C correctly expresses this as lim⁡h→0C(50+h)−C(50)h\lim_{h\to 0}\frac{C(50+h)-C(50)}{h}limh→0​hC(50+h)−C(50)​, capturing the exact rate of cost increase per additional unit at that production level. Choices A and D calculate average costs over intervals, while choices B and E approximate the marginal cost using one-unit differences but without the limiting process. A common misconception is that C(51)−C(50)C(51)-C(50)C(51)−C(50) gives the instantaneous marginal cost, but this is actually the average cost of producing the 51st unit. Remember: instantaneous rates in economics, like marginal cost, require derivatives (limits) to capture the rate at a precise production level.

Question 20

A population is modeled by P(t)P(t)P(t). Which expression represents the instantaneous growth rate at t=6t=6t=6?

  1. P(6)−P(5)6−5\dfrac{P(6)-P(5)}{6-5}6−5P(6)−P(5)​
  2. P(6)−P(0)P(6)-P(0)P(6)−P(0)
  3. P(8)−P(4)8−4\dfrac{P(8)-P(4)}{8-4}8−4P(8)−P(4)​
  4. lim⁡h→0P(6+h)−P(6)h\displaystyle \lim_{h\to 0}\dfrac{P(6+h)-P(6)}{h}h→0lim​hP(6+h)−P(6)​ (correct answer)
  5. P(7)−P(6)7−6\dfrac{P(7)-P(6)}{7-6}7−6P(7)−P(6)​

Explanation: The instantaneous growth rate requires finding how fast the population is changing at exactly t=6, which means we need the derivative. Options A and E calculate average growth rates over the intervals [5,6] and [6,7] respectively, while option C gives the average rate from t=4 to t=8. Option B provides the total population change from t=0 to t=6, which isn't a rate. Only option D uses the limit definition, examining what happens as we consider smaller and smaller time intervals around t=6. This limiting process is essential for capturing the instantaneous rate. A helpful strategy: if you see "instantaneous," look for the limit notation—average rates never use limits.