AP Calculus AB Quiz: Mean Value Theorem

What this quiz covers

This quiz focuses on Mean Value Theorem, 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.

Question 1

1. Yes, because $t$ is continuous on $[0,1]$ and differentiable on $(0,1)$, so some $c$ has $t'(c)=\dfrac{5-0}{1-0}=5$. (correct answer)
2. No, because MVT does not apply on intervals of length 1.
3. Yes, because $t(1)=5$.
4. No, because MVT requires $t(0)=t(1)$.
5. Yes, because $t$ is continuous so its derivative is 5 everywhere.

Explanation: The Mean Value Theorem requires t to be continuous on [0,1] and differentiable on (0,1). Since these conditions are satisfied, MVT guarantees there exists c in (0,1) where t'(c) equals the average rate of change. The average rate of change is (t(1)-t(0))/(1-0) = (5-0)/1 = 5/1 = 5. Therefore, MVT does guarantee a point where t'(c) = 5. A common error is thinking MVT doesn't apply to intervals of length 1, but the theorem works on any interval where the function meets the required conditions. Interval length doesn't affect MVT's validity.

Question 2

1. Yes, because $f$ is continuous on $[1,13]$ and differentiable on $(1,13)$, so some $c$ has $f'(c)=\dfrac{17-5}{13-1}=1$. (correct answer)
2. No, because MVT requires $f(1)=f(13)$.
3. Yes, because $f$ is continuous so $f'(c)=1$ at the midpoint.
4. Yes, because $f(13)-f(1)=12$.
5. No, because the average rate of change is 12.

Explanation: The Mean Value Theorem applies when f is continuous on [1,13] and differentiable on (1,13). Since both conditions are met, MVT guarantees there exists c in (1,13) where f'(c) equals the average rate of change. The average rate of change is (f(13)-f(1))/(13-1) = (17-5)/12 = 12/12 = 1. Therefore, MVT does guarantee a point where f'(c) = 1. Students sometimes confuse the total change in function values (12) with the average rate of change (1), but MVT specifically guarantees the derivative equals the average rate of change, which requires dividing by the interval length.

Question 3

1. Yes, because $g$ changes from 7 to 1, so $g'(c)=-1$ somewhere.
2. Yes, because $g$ is continuous on $[-2,4]$ and differentiable on $(-2,4)$, so some $c$ has $g'(c)=\dfrac{1-7}{4-(-2)}=-1$. (correct answer)
3. No, because $g(-2)\neq g(4)$, so the Mean Value Theorem cannot apply.
4. Yes, because $g$ is differentiable on $[-2,4]$, so $g'(c)$ equals $g(4)-g(-2)$.
5. No, because the Mean Value Theorem requires $g$ to be increasing.

Explanation: The Mean Value Theorem requires continuity on [a,b] and differentiability on (a,b). Given that g satisfies these conditions on [-2,4], MVT guarantees there exists c in (-2,4) where g'(c) equals the average rate of change. Computing: (g(4)-g(-2))/(4-(-2)) = (1-7)/6 = -1. Therefore, MVT does guarantee a point where g'(c) = -1. Students sometimes mistakenly think MVT requires equal endpoint values, but that's Rolle's Theorem. MVT works for any function meeting the continuity and differentiability requirements, regardless of whether endpoints are equal. Verify both conditions before concluding MVT applies.

Question 4

1. Yes, because $f(2)=f(8)$, so the average rate of change is 0 and MVT gives $f'(c)=0$ for some $c$. (correct answer)
2. Yes, because equal endpoint values guarantee $f$ has a minimum at the midpoint.
3. No, because $f$ might be constant only at the endpoints.
4. Yes, because $f$ is continuous, so $f'(c)=0$ must occur.
5. No, because MVT requires $f(2)\ne f(8)$.

Explanation: The Mean Value Theorem applies when f is continuous on [2,8] and differentiable on (2,8). Since both conditions are met, MVT guarantees there exists c in (2,8) where f'(c) equals the average rate of change. The average rate of change is (f(8)-f(2))/(8-2) = (5-5)/6 = 0. Therefore, MVT does guarantee a point where f'(c) = 0. This is actually a special case leading toward Rolle's Theorem - when endpoints are equal, the average rate of change is zero, so MVT guarantees at least one point with zero derivative. Students should recognize that equal endpoints make MVT application straightforward and powerful.

Question 5

1. Yes, because $G$ is continuous on $[2,6]$ and differentiable on $(2,6)$, so some $c$ has $G'(c)=\dfrac{7-(-1)}{6-2}=2$. (correct answer)
2. No, because $G(2)$ is negative.
3. Yes, because $G$ crosses 0.
4. No, because MVT requires $G(2)=G(6)$.
5. Yes, because $G$ is continuous, so the derivative must be constant.

Explanation: The Mean Value Theorem requires G to be continuous on [2,6] and differentiable on (2,6). Since these conditions are satisfied, MVT guarantees there exists c in (2,6) where G'(c) equals the average rate of change. Computing: (G(6)-G(2))/(6-2) = (7-(-1))/4 = 8/4 = 2. Therefore, MVT does guarantee a point where G'(c) = 2. A common error is thinking negative function values prevent MVT application, but the theorem works regardless of whether function values are positive, negative, or mixed. The key is meeting the continuity and differentiability requirements, not the signs of the function values.

Question 6

1. Yes, because $S$ is continuous on $[-6,-2]$ and differentiable on $(-6,-2)$, so some $c$ has $S'(c)=\dfrac{4-(-4)}{-2-(-6)}=2$. (correct answer)
2. No, because negative inputs prevent using MVT.
3. Yes, because $S$ changes sign.
4. No, because MVT requires $S(-6)=S(-2)$.
5. Yes, because $S$ is continuous so its derivative must be 2 at the midpoint.

Explanation: The Mean Value Theorem requires S to be continuous on [-6,-2] and differentiable on (-6,-2). Since these conditions are met, MVT guarantees there exists c in (-6,-2) where S'(c) equals the average rate of change. The average rate of change is (S(-2)-S(-6))/(-2-(-6)) = (4-(-4))/4 = 8/4 = 2. Therefore, MVT does guarantee a point where S'(c) = 2. A common misconception is that negative input values prevent MVT application, but the theorem works on any interval where the function satisfies the required conditions. The sign or location of the x-values doesn't affect MVT's validity.

Question 7

1. Yes, because $R$ is continuous on $[0,8]$ and differentiable on $(0,8)$, so some $c$ has $R'(c)=\dfrac{18-2}{8-0}=2$. (correct answer)
2. No, because MVT only works if $R$ is decreasing.
3. Yes, because $R(8)-R(0)=16$.
4. Yes, because $R$ is continuous, so $R'(c)=2$ for all $c$.
5. No, because the derivative might not exist at endpoints.

Explanation: The Mean Value Theorem applies when R is continuous on [0,8] and differentiable on (0,8). Since both conditions are satisfied, MVT guarantees there exists c in (0,8) where R'(c) equals the average rate of change. The average rate of change is (R(8)-R(0))/(8-0) = (18-2)/8 = 16/8 = 2. Therefore, MVT does guarantee a point where R'(c) = 2. Students might incorrectly think MVT only works for decreasing functions, but the theorem applies to increasing, decreasing, and mixed-behavior functions alike. The direction of change doesn't restrict MVT's validity - only the continuity and differentiability conditions matter.

Question 8

1. Yes, because $q$ is continuous on $[4,6]$ and differentiable on $(4,6)$, so some $c$ has $q'(c)=\dfrac{9-1}{6-4}=4$. (correct answer)
2. No, because MVT requires $q(4)=q(6)$.
3. Yes, because $q(6)-q(4)=8$.
4. Yes, because $q$ is continuous, so its derivative must be 4 at $x=5$.
5. No, because the derivative cannot exceed 1 on a short interval.

Explanation: The Mean Value Theorem applies when q is continuous on [4,6] and differentiable on (4,6). Since both conditions are satisfied, MVT guarantees there exists c in (4,6) where q'(c) equals the average rate of change. The average rate of change is (q(6)-q(4))/(6-4) = (9-1)/2 = 8/2 = 4. Therefore, MVT does guarantee a point where q'(c) = 4. Students might incorrectly think derivatives cannot exceed 1 on short intervals, but MVT places no restrictions on the magnitude of the guaranteed derivative. The derivative can be any value that equals the computed average rate of change.

Question 9

1. Yes, because $n$ is continuous on $[-4,0]$ and differentiable on $(-4,0)$, so some $c$ has $n'(c)=\dfrac{10-2}{0-(-4)}=2$. (correct answer)
2. No, because the interval ends at 0.
3. Yes, because $n(0)-n(-4)=8$.
4. No, because MVT requires $n(-4)=n(0)$.
5. Yes, because $n$ is continuous, so it has slope 2 everywhere.

Explanation: The Mean Value Theorem requires n to be continuous on [-4,0] and differentiable on (-4,0). Since these conditions are satisfied, MVT guarantees there exists c in (-4,0) where n'(c) equals the average rate of change. Computing: (n(0)-n(-4))/(0-(-4)) = (10-2)/4 = 8/4 = 2. Therefore, MVT does guarantee a point where n'(c) = 2. A common misconception is that intervals ending at zero create special restrictions, but MVT applies to any interval where the function satisfies the required conditions. The specific endpoints don't affect the theorem's validity.

Question 10

1. Yes, because $T$ is continuous on $[2,5]$ and differentiable on $(2,5)$, so some $c$ has $T'(c)=\dfrac{7-1}{5-2}=2$. (correct answer)
2. Yes, because $T(5)-T(2)=6$.
3. No, because MVT needs $T$ to be linear.
4. Yes, because $T$ is continuous, so $T'(c)=2$ must occur.
5. No, because the interval does not include 0.

Explanation: The Mean Value Theorem applies when T is continuous on [2,5] and differentiable on (2,5). Since both conditions are satisfied, MVT guarantees there exists c in (2,5) where T'(c) equals the average rate of change. The average rate of change is (T(5)-T(2))/(5-2) = (7-1)/3 = 6/3 = 2. Therefore, MVT does guarantee a point where T'(c) = 2. Students sometimes think MVT requires linear functions, but the theorem applies to any function meeting the continuity and differentiability requirements. The function can have any shape - curved, linear, or mixed - as long as it satisfies the basic conditions.

Question 11

1. No, because MVT requires $g$ to be differentiable on $[2,10]$, including endpoints.
2. Yes, because $g$ is continuous on $[2,10]$ and differentiable on $(2,10)$, so some $c$ satisfies $g'(c)=\tfrac{1}{2}$. (correct answer)
3. Yes, because $g(10)>g(2)$, so there must be $c$ with $g'(c)=0$.
4. Yes, because $g$ is continuous on $(2,10)$, so a tangent line parallel to the secant must occur.
5. No, because MVT can only be applied if $g(2)=g(10)$.

Explanation: The function g meets both MVT conditions: continuous on [2,10] and differentiable on (2,10). The average rate of change is (5-1)/(10-2) = 4/8 = 1/2. MVT guarantees there exists c in (2,10) where g'(c) = 1/2. Choice B correctly identifies the conditions and the derivative value. A common mistake is thinking MVT requires differentiability at endpoints - it only needs differentiability on the open interval. Strategy: Remember MVT needs differentiability only on the open interval (a,b), not at the endpoints.

Question 12

1. Yes, because $p$ crosses the $x$-axis, so there is $c$ with $p'(c)=-1$.
2. No, because MVT applies only if $p$ is increasing.
3. Yes, because $p$ is continuous on $[-3,1]$ and differentiable on $(-3,1)$, so some $c$ satisfies $p'(c)=-1$. (correct answer)
4. No, because continuity on $[-3,1]$ is unnecessary and may fail.
5. Yes, because $p(1)=0$ guarantees a point where the derivative equals $0$.

Explanation: Function p satisfies both MVT requirements: continuous on [-3,1] and differentiable on (-3,1). The average rate of change is (0-4)/(1-(-3)) = -4/4 = -1. MVT guarantees there exists c in (-3,1) where p'(c) = -1. Choice C correctly states the conditions and conclusion. A common confusion is focusing on the function crossing the x-axis (p(1) = 0), which relates to IVT for finding roots, not MVT for finding derivative values. Strategy: Don't be distracted by special function values; MVT only depends on endpoint values and the two key conditions.

Question 13

1. No, because $p(0)$ and $p(3)$ have opposite signs, which only implies a root by IVT.
2. Yes, because $p$ is differentiable on $[0,3]$, so there is $c$ with $p'(c)=2$.
3. Yes, because $p$ is continuous on $[0,3]$ and differentiable on $(0,3)$, so some $c$ satisfies $p'(c)=2$. (correct answer)
4. No, because MVT requires $p(0)=p(3)$.
5. Yes, because continuity alone guarantees a point where the derivative equals the average slope.

Explanation: The function p satisfies both MVT conditions: continuous on [0,3] and differentiable on (0,3). The average rate of change is (2-(-4))/(3-0) = 6/3 = 2. Therefore, MVT guarantees there exists c in (0,3) where p'(c) = 2. Choice C correctly identifies both conditions and the correct derivative value. A common confusion is mixing up MVT with IVT - opposite signs at endpoints relates to finding roots (IVT), not to finding where derivatives equal average slopes (MVT). Strategy: Don't let function values crossing zero distract you; MVT only cares about continuity, differentiability, and average rate of change.

Question 14

1. Yes, because $s$ is continuous on $[0,8]$, differentiable on $(0,8)$, and $\frac{s(8)-s(0)}{8-0}=\frac{1-9}{8}=-1$. (correct answer)
2. Yes, because $-1$ is between $s(0)$ and $s(8)$.
3. No, because $s(0)>s(8)$ implies $s'(x)$ is always negative but not necessarily $-1$.
4. Yes, because $s$ is differentiable on $(0,8)$, so it must match the average slope.
5. No, because MVT does not apply on closed intervals.

Explanation: The function s meets both MVT conditions: continuous on [0,8] and differentiable on (0,8). The average rate of change is (s(8)-s(0))/(8-0) = (1-9)/8 = -8/8 = -1, which is exactly the derivative value in question. Therefore, MVT guarantees the existence of at least one c in (0,8) where s'(c) = -1. While s is decreasing overall (since s(8) < s(0)), this doesn't mean s'(x) is constantly -1; the derivative could vary while remaining negative. A common mistake is thinking MVT doesn't apply to closed intervals, but it specifically requires continuity on the closed interval [a,b]. Remember: MVT connects average and instantaneous rates of change, regardless of whether the function increases or decreases.

Question 15

1. Yes, because $w$ is continuous on $[-1,1]$ and differentiable on $(-1,1)$, so some $c$ has $w'(c)=\dfrac{6-2}{1-(-1)}=2$. (correct answer)
2. No, because the interval length is 2.
3. Yes, because $w(1)$ is 6.
4. No, because $w(-1)$ is not 0.
5. Yes, because $w$ is continuous, so $w'(c)=2$ must occur.

Explanation: The Mean Value Theorem requires w to be continuous on [-1,1] and differentiable on (-1,1). Since these conditions are met, MVT guarantees there exists c in (-1,1) where w'(c) equals the average rate of change. The average rate of change is (w(1)-w(-1))/(1-(-1)) = (6-2)/2 = 4/2 = 2. Therefore, MVT does guarantee a point where w'(c) = 2. Students might think interval length matters for MVT application, but the theorem works on any interval where the function satisfies the required conditions. Whether the interval is long or short doesn't affect MVT's validity - only continuity and differentiability matter.

Question 16

1. No, because MVT requires $h(0)=h(9)$.
2. Yes, because $h$ is continuous, so $h'(c)=1$ must occur.
3. Yes, because $h$ is continuous on $[0,9]$ and differentiable on $(0,9)$, so some $c$ has $h'(c)=\dfrac{11-2}{9-0}=1$. (correct answer)
4. No, because the average rate of change is 9.
5. Yes, because $h(9)-h(0)=9$.

Explanation: The Mean Value Theorem applies when h is continuous on [0,9] and differentiable on (0,9). Since both conditions are satisfied, MVT guarantees there exists c in (0,9) where h'(c) equals the average rate of change. The average rate of change is (h(9)-h(0))/(9-0) = (11-2)/9 = 9/9 = 1. Therefore, MVT does guarantee a point where h'(c) = 1. Students sometimes confuse the total change in function values (9) with the average rate of change (1), but MVT specifically provides the derivative equal to the average rate of change, not the total change.

Question 17

1. Yes, because $u(9)-u(1)=16$, so $u'(c)=16$ for some $c$.
2. Yes, because $u$ is continuous on $[1,9]$ and differentiable on $(1,9)$, so some $c$ has $u'(c)=\dfrac{16-0}{9-1}=2$. (correct answer)
3. No, because MVT requires $u(1)=u(9)$.
4. Yes, because $u$ is continuous, so its derivative must take all values between 0 and 16.
5. No, because the derivative might not exist anywhere.

Explanation: The Mean Value Theorem requires u to be continuous on [1,9] and differentiable on (1,9). Since these conditions are satisfied, MVT guarantees there exists c in (1,9) where u'(c) equals the average rate of change. Computing: (u(9)-u(1))/(9-1) = (16-0)/8 = 16/8 = 2. Therefore, MVT does guarantee a point where u'(c) = 2. Students sometimes confuse the change in function values u(9)-u(1) = 16 with the derivative value, but MVT specifically guarantees the derivative equals the average rate of change, which requires dividing by the interval length. Always compute (f(b)-f(a))/(b-a) for MVT applications.

Question 18

1. Yes, because $d$ is continuous on $[1,2]$ and differentiable on $(1,2)$, so some $c$ has $d'(c)=\dfrac{-1-(-5)}{2-1}=4$. (correct answer)
2. No, because both endpoint values are negative.
3. Yes, because $d$ increases by 4.
4. No, because MVT requires $d(1)=d(2)$.
5. Yes, because $d$ is continuous, so the derivative equals 4 everywhere.

Explanation: The Mean Value Theorem applies when d is continuous on [1,2] and differentiable on (1,2). Since both conditions are satisfied, MVT guarantees there exists c in (1,2) where d'(c) equals the average rate of change. The average rate of change is (d(2)-d(1))/(2-1) = (-1-(-5))/1 = 4/1 = 4. Therefore, MVT does guarantee a point where d'(c) = 4. A common misconception is that negative function values prevent MVT application, but the theorem works regardless of whether function values are positive, negative, or mixed. Only the continuity and differentiability conditions matter.

Question 19

1. Yes, because $N$ is continuous on $[-2,2]$ and differentiable on $(-2,2)$, so some $c$ has $N'(c)=\dfrac{-3-5}{2-(-2)}=-2$. (correct answer)
2. Yes, because $N$ changes sign, so the derivative is $-2$.
3. No, because MVT cannot be used when values are negative.
4. No, because the average rate of change is $-8$.
5. Yes, because $N$ is continuous so $N'(c)$ takes all values between 5 and $-3$.

Explanation: The Mean Value Theorem requires N to be continuous on [-2,2] and differentiable on (-2,2). Since these conditions are satisfied, MVT guarantees there exists c in (-2,2) where N'(c) equals the average rate of change. Computing: (N(2)-N(-2))/(2-(-2)) = (-3-5)/4 = -8/4 = -2. Therefore, MVT does guarantee a point where N'(c) = -2. A common misconception is that negative function values prevent MVT application, but the theorem works with any real-valued function meeting the required conditions. The signs of the function values don't affect MVT's validity.

Question 20

1. Yes, because $r$ is continuous on $[-3,1]$ and differentiable on $(-3,1)$, so some $c$ has $r'(c)=\dfrac{8-0}{1-(-3)}=2$. (correct answer)
2. No, because $r(-3)=0$ means $r'(c)=0$ somewhere instead.
3. Yes, because $r(1)=8$ is greater than $r(-3)=0$.
4. No, because MVT only applies when the interval is symmetric about 0.
5. Yes, because the function crosses $y=2$.

Explanation: The Mean Value Theorem requires r to be continuous on [-3,1] and differentiable on (-3,1). Since these conditions are satisfied, MVT guarantees there exists c in (-3,1) where r'(c) equals the average rate of change. Computing: (r(1)-r(-3))/(1-(-3)) = (8-0)/4 = 8/4 = 2. Therefore, MVT does guarantee a point where r'(c) = 2. A common misconception is that having one endpoint equal to zero creates special restrictions, but zero is just another function value. MVT applies regardless of the specific numerical values at the endpoints, as long as the continuity and differentiability conditions are met.