AP Calculus BC 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 BC.

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 $p$ is continuous, a slope of $-1$ must occur.
2. No; MVT cannot be applied when the interval crosses $0$.
3. Yes; $\frac{1-9}{6-(-2)}=-1$ and hypotheses satisfy MVT. (correct answer)
4. No; MVT requires $p(-2)=p(6)$.
5. Yes; differentiability on $(-2,6)$ alone guarantees it.

Explanation: The MVT ensures that the derivative equals the secant slope at some interior point if continuity holds on [a, b] and differentiability on (a, b). For p, these are met, with average rate (1 - 9) / (6 - (-2)) = -1, so a c exists with p'(c) = -1. People often mistakenly require equal endpoints for MVT, but that's only for Rolle's zero-slope case. Another error is assuming the interval can't cross zero, but MVT has no such restriction. Choice C properly calculates and confirms conditions. To transfer this, always compute [f(b) - f(a)] / (b - a) after verifying hypotheses, regardless of interval signs.

Question 2

1. No; MVT cannot be applied when the left endpoint is negative.
2. Yes; because $h$ is differentiable on $(-3,1)$, it must have slope $2$ somewhere.
3. Yes; $\frac{6-(-2)}{1-(-3)}=2$ and hypotheses match MVT. (correct answer)
4. No; MVT requires $h(-3)=h(1)$.
5. Yes; continuity on $[-3,1]$ alone guarantees it.

Explanation: MVT works across intervals including negatives, provided the hypotheses hold. For h continuous on [-3, 1] and differentiable on (-3, 1), h(-3) = -2 and h(1) = 6, the slope is (6 - (-2)) / (1 - (-3)) = 2, guaranteeing h'(c) = 2. The theorem applies fully. A common misuse is assuming negative endpoints prevent application. Requiring equal endpoints is a Rolle's confusion. Choice C accurately reflects this. A transferable strategy is to compute carefully with negative values and verify conditions independently.

Question 3

1. No; the interval length is too small for MVT to apply.
2. Yes; since $\frac{4-0}{2-1}=4$ and MVT conditions are satisfied, such $c$ exists. (correct answer)
3. Yes; because $f$ is continuous on $(1,2)$, the derivative must equal $4$ somewhere.
4. No; MVT requires $f(1)=f(2)$.
5. Yes; any differentiable function has a point where $f'(c)$ equals $f(2)$.

Explanation: Even on short intervals, MVT applies if conditions are met, ensuring the derivative equals the secant slope somewhere. For f continuous on [1, 2] and differentiable on (1, 2), with f(1) = 0 and f(2) = 4, the rate is (4 - 0) / (2 - 1) = 4, guaranteeing f'(c) = 4. The small length does not invalidate it. A common misuse is believing interval length affects applicability, but it doesn't. Requiring equal endpoints is another confusion with Rolle's. Choice B correctly states this. A transferable strategy is to ignore interval length and focus on hypotheses and computation.

Question 4

1. No; MVT requires $q(2)=q(8)$.
2. Yes; because $q$ is continuous on $[2,8]$ and differentiable on $(2,8)$, some $c$ has $q'(c)=\dfrac{1-7}{8-2}=-1$. (correct answer)
3. Yes; continuity on $[2,8]$ alone guarantees a point where $q'(c)=-1$.
4. No; MVT requires $q$ to be differentiable at $x=2$ and $x=8$.
5. Yes; since $q(2)>q(8)$, there must be $c$ with $q'(c)=-1$ even if $q$ is not differentiable.

Explanation: The function q satisfies both MVT conditions: continuous on [2,8] and differentiable on (2,8). The average rate of change over the interval is (q(8)-q(2))/(8-2) = (1-7)/6 = -6/6 = -1. Therefore, MVT guarantees the existence of at least one c in (2,8) where q'(c) = -1. A common error is thinking MVT requires the function to be increasing or that it cannot guarantee negative derivative values. The theorem works regardless of whether the function increases or decreases. Always check the two conditions and calculate the average rate of change to determine what f'(c) must equal.

Question 5

1. Yes, because $q(2)=q(6)$ and $q$ is continuous on $[2,6]$ and differentiable on $(2,6)$. (correct answer)
2. No, because $q$ might not have a maximum value on $[2,6]$.
3. Yes, because $q$ is differentiable at $x=2$ and $x=6$.
4. Yes, because $q(2)=q(6)$ implies $q$ is constant on $[2,6]$.
5. No, because the Mean Value Theorem requires $q(2)\ne q(6)$.

Explanation: Since q is continuous on [2,6] and differentiable on (2,6), MVT applies. The average rate of change is (q(6)-q(2))/(6-2) = (9-9)/4 = 0/4 = 0. Therefore, MVT guarantees there exists at least one c in (2,6) where q'(c) = 0. This is actually a special case of MVT known as Rolle's Theorem, which applies when the function values at the endpoints are equal. A common error is thinking equal endpoints mean the function is constant throughout—q could rise and fall between the endpoints. The existence of a maximum on [2,6] isn't required by MVT. Remember: When f(a) = f(b), MVT guarantees a horizontal tangent somewhere in (a,b).

Question 6

1. Yes, because $t$ is continuous on $[-4,-1]$ and differentiable on $(-4,-1)$. (correct answer)
2. No, because $t$ must cross the $x$-axis on $[-4,-1]$.
3. Yes, because $t(-4)$ and $t(-1)$ are both positive.
4. No, because the Mean Value Theorem requires $t$ to be differentiable at $x=-4$ and $x=-1$.
5. Yes, because $t$ has an average rate of change of $-1$ and is differentiable somewhere.

Explanation: The function t satisfies MVT's conditions: continuous on [-4,-1] and differentiable on (-4,-1). The average rate of change is (t(-1)-t(-4))/((-1)-(-4)) = (2-5)/3 = -3/3 = -1. Therefore, MVT guarantees there exists at least one c in (-4,-1) where t'(c) = -1. Students might incorrectly think MVT requires the function to cross the x-axis or that both endpoint values being positive matters—neither affects MVT's applicability. MVT doesn't require differentiability at the endpoints, only on the open interval. Remember: MVT is about connecting average and instantaneous rates of change, not about specific function behaviors or zero crossings.

Question 7

1. Yes; because $u$ is differentiable on $[-4,0]$, MVT gives $u'(c)=2$.
2. No; MVT requires $u(-4)=u(0)$.
3. Yes; continuity on $(-4,0)$ is sufficient to guarantee $u'(c)=2$.
4. Yes; $u$ is continuous on $[-4,0]$ and differentiable on $(-4,0)$, so some $c$ satisfies $u'(c)=\dfrac{7-(-1)}{0-(-4)}=2$. (correct answer)
5. No; MVT cannot be used when an endpoint is $0$.

Explanation: The function u meets MVT's hypotheses: continuous on [-4,0] and differentiable on (-4,0). The average rate of change is [u(0)-u(-4)]/(0-(-4)) = [7-(-1)]/4 = 8/4 = 2. Therefore, MVT guarantees there exists some c in (-4,0) where u'(c) = 2. A common error is thinking MVT cannot be used when an endpoint is 0—this is false. MVT applies to any interval [a,b] with a < b, including intervals ending at 0. The theorem doesn't care about the specific values of the endpoints, only that the function is continuous and differentiable. Strategy: Focus on verifying the conditions, not the particular numbers involved.

Question 8

1. Yes; since $t$ is continuous on $[-4,2]$ and differentiable on $(-4,2)$, some $c$ has $t'(c)=\dfrac{4-10}{2-(-4)}=-1$. (correct answer)
2. No; MVT applies only when the function is linear.
3. Yes; because $t(-4)>t(2)$, there must be $c$ with $t'(c)=-1$ even without continuity.
4. No; MVT requires $t$ to be differentiable at $x=-4$ and $x=2$.
5. Yes; continuity on $(-4,2)$ alone guarantees a point where $t'(c)=-1$.

Explanation: The function t satisfies MVT's hypotheses: continuous on [-4,2] and differentiable on (-4,2). The average rate of change is (t(2)-t(-4))/(2-(-4)) = (4-10)/6 = -6/6 = -1. By MVT, there must exist at least one c in (-4,2) where t'(c) = -1. A common error is thinking MVT only applies to linear functions or that it requires differentiability at the endpoints. The theorem works for any function meeting the continuity and differentiability conditions, regardless of its shape. Always verify the conditions first, then calculate the average rate of change to find the guaranteed derivative value.

Question 9

1. No; MVT requires $f$ to be differentiable on $[7,9]$.
2. Yes; because $f$ is continuous on $(7,9)$, it must have derivative $4$.
3. Yes; $\frac{11-3}{9-7}=4$ and MVT hypotheses are satisfied. (correct answer)
4. No; MVT applies only when $f(7)=f(9)$.
5. Yes; endpoint values guarantee a point with derivative equal to the larger endpoint value.

Explanation: MVT applies when a function is continuous on the closed interval and differentiable on the open one, guaranteeing a match between derivative and average slope. Here, f satisfies this on [7, 9] and (7, 9), with slope (11 - 3) / (9 - 7) = 4, so yes to f'(c) = 4. A typical misuse is thinking MVT needs differentiability at endpoints, but it does not. Confusing it with requiring equal endpoints is another common mistake, as that's Rolle's. Choice C correctly identifies the details. Strategically, verify conditions first, then calculate the average to apply MVT confidently in various scenarios.

Question 10

1. Yes; because $p$ is continuous, it must have a point with slope $-2$.
2. No; MVT requires $p$ to be decreasing on the entire interval.
3. Yes; $\frac{1-5}{2-0}=-2$ and the hypotheses satisfy MVT, so such $c$ exists. (correct answer)
4. No; MVT requires $p(0)=p(2)$.
5. Yes; differentiability on $[0,2]$ alone guarantees it.

Explanation: MVT ensures a matching derivative for the computed average rate when hypotheses hold. For p continuous on [0, 2] and differentiable on (0, 2), p(0) = 5 and p(2) = 1, the slope is (1 - 5) / (2 - 0) = -2, guaranteeing p'(c) = -2. The application is correct. A common misuse is requiring the function to be decreasing everywhere. Thinking equal endpoints are needed is a Rolle's error. Choice C accurately states this. A transferable strategy is to check for monotonicity assumptions and avoid them in MVT applications.

Question 11

1. Yes; since $\frac{15-0}{6-1}=3$ and MVT hypotheses hold. (correct answer)
2. No; MVT requires $g$ to be increasing everywhere.
3. Yes; differentiability on $(1,6)$ alone guarantees such a $c$.
4. No; MVT requires $g(1)=g(6)$.
5. Yes; continuity on $[1,6]$ alone guarantees a point where $g'(c)=3$.

Explanation: The theorem promises a point where f'(c) equals the overall average change rate, given continuity on [a, b] and differentiability on (a, b). For g, conditions hold, average is (15 - 0) / (6 - 1) = 3, guaranteeing g'(c) = 3. Misuse often involves thinking MVT requires the function to be increasing, but it applies to any behavior as long as hypotheses are met. Believing equal endpoints are needed is incorrect outside of Rolle's. Choice A is accurate. A transferable approach: Check hypotheses, compute slope, and recall MVT doesn't impose monotonicity.

Question 12

1. Yes; $\frac{8-(-1)}{4-1}=3$ and MVT conditions are satisfied. (correct answer)
2. No; MVT requires $h(1)=h(4)$.
3. Yes; because $h$ is differentiable on $(1,4)$, it must take slope $3$.
4. No; continuity on $(1,4)$ is required instead of $[1,4]$.
5. Yes; any continuous function has a point where derivative equals $3$.

Explanation: MVT links secant and tangent slopes for functions meeting the criteria. For h continuous on [1, 4] and differentiable on (1, 4), h(1) = -1 and h(4) = 8, the average is (8 - (-1)) / (4 - 1) = 3, guaranteeing h'(c) = 3. The theorem applies. A common misuse is swapping continuity and differentiability intervals. Requiring equal endpoints is a mistake. Choice A accurately reflects the reasoning. A transferable strategy is to double-check interval types when applying hypotheses.

Question 13

1. Yes; because $p$ is continuous, it must have derivative $2$ somewhere.
2. Yes; $\frac{19-3}{7-(-1)}=2$ and MVT hypotheses are met. (correct answer)
3. No; MVT requires $p(-1)=p(7)$.
4. No; MVT requires differentiability on $[-1,7]$.
5. Yes; differentiability on $(-1,7)$ alone guarantees it.

Explanation: Under MVT, derivative matches average rate somewhere in the open interval if conditions are satisfied. p does, with (19 - 3) / (7 - (-1)) = 2, guaranteeing the c. Misuse: requiring differentiability on closed interval, but not needed. Thinking equal endpoints are required confuses with Rolle's. Choice B verifies correctly. Always differentiate between closed continuity and open differentiability in checks.

Question 14

1. Yes; MVT guarantees $g'(c)=\frac{0-9}{3-0}=-3$ for some $c$. (correct answer)
2. No; MVT requires $g$ to be differentiable at $0$ and $3$.
3. Yes; because $g$ is continuous, it must have derivative $-3$ somewhere.
4. No; MVT applies only if $g$ is linear.
5. Yes; Rolle’s Theorem applies since $g(0)\neq g(3)$.

Explanation: MVT connects overall change to local rates for continuous and differentiable functions as specified. Given g continuous on [0, 3] and differentiable on (0, 3), g(0) = 9 and g(3) = 0, the average is (0 - 9) / (3 - 0) = -3, so g'(c) = -3 is guaranteed. Conditions are satisfied. A common misuse is thinking MVT requires differentiability at endpoints. Invoking Rolle's incorrectly when endpoints differ is another error. Choice A properly applies MVT. A transferable strategy is to distinguish MVT from special cases like Rolle's by checking endpoint equality.

Question 15

1. No; MVT requires $f$ to be decreasing everywhere.
2. Yes; $\frac{2-12}{5-0}=-2$ and the hypotheses satisfy MVT. (correct answer)
3. Yes; because $f$ is differentiable, it must take slope $-2$.
4. No; MVT applies only when $f(0)=f(5)$.
5. Yes; continuity on $(0,5)$ alone guarantees it.

Explanation: MVT guarantees f'(c) = [f(b) - f(a)] / (b - a) under the given conditions. For f, yes, with -2 average, so guaranteed. Common misuse: assuming decreasing function required, but not. Equal endpoints not needed. Choice B is right. Strategy: Ignore function behavior assumptions beyond hypotheses.

Question 16

1. Yes; because $f$ is differentiable on $(-3,3)$, it must have derivative $2$.
2. Yes; $\frac{12-0}{3-(-3)}=2$ and MVT hypotheses are met. (correct answer)
3. No; MVT requires $f(-3)=f(3)$.
4. No; MVT requires $f$ to be increasing everywhere.
5. Yes; continuity on $[-3,3]$ alone guarantees it.

Explanation: The Mean Value Theorem (MVT) states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in (a, b) where the derivative f'(c) equals the average rate of change over [a, b], calculated as [f(b) - f(a)] / (b - a). In this case, f is continuous on [-3, 3] and differentiable on (-3, 3), satisfying the hypotheses, and the average rate is (12 - 0) / (3 - (-3)) = 2, so MVT guarantees a c where f'(c) = 2. A common misuse is confusing MVT with Rolle's Theorem, which requires f(a) = f(b) for a zero derivative somewhere, but MVT applies regardless of whether endpoints are equal. Another error is thinking continuity alone suffices, but differentiability on the open interval is crucial for the existence of the derivative. Choice B correctly identifies the average rate and confirms the hypotheses are met. To apply MVT effectively, always verify both continuity on the closed interval and differentiability on the open interval before computing the average rate of change.

Question 17

1. No; MVT cannot be used when function values are negative.
2. Yes; because $f$ is differentiable on $(-1,3)$, some $c$ has $f'(c)=-2$.
3. Yes; the average slope is $\frac{-4-4}{3-(-1)}=-2$, and conditions match MVT. (correct answer)
4. No; MVT requires $f$ to be continuous on $(-1,3)$ only.
5. No; MVT requires $f(-1)=f(3)$.

Explanation: MVT ensures that the instantaneous rate equals the average rate at some point if continuity and differentiability conditions are fulfilled. For f continuous on [-1, 3] and differentiable on (-1, 3), with f(-1) = 4 and f(3) = -4, the average is (-4 - 4) / (3 - (-1)) = -2, guaranteeing f'(c) = -2. The theorem applies regardless of negative values, as conditions hold. A common misuse is thinking MVT fails for negative function values or slopes, which is incorrect. Another error is requiring equal endpoints, but that's for Rolle's. Choice C properly states the calculation and conditions. A transferable strategy is to compute the secant slope inclusively and confirm MVT hypotheses to avoid misapplication.

Question 18

1. Yes; since $g(0)\neq g(4)$, Rolle’s Theorem guarantees $g'(c)=1$.
2. No; MVT requires $g$ to be differentiable at $0$ and $4$.
3. Yes; MVT guarantees $g'(c)=\frac{5-1}{4-0}=1$ under the stated conditions. (correct answer)
4. No; continuity on $[0,4]$ alone cannot guarantee any derivative value.
5. Yes; any function with endpoints given has a point where derivative equals endpoint slope.

Explanation: The core of MVT is linking average and instantaneous rates via specific conditions on an interval. Given g continuous on $[0, 4]$ and differentiable on $(0, 4)$, with $g(0) = 1$ and $g(4) = 5$, the average rate is $\frac{5-1}{4-0} = 1$, so MVT guarantees $g'(c) = 1$. All requirements are satisfied, enabling the conclusion. A frequent misuse is demanding differentiability at endpoints, which MVT does not require. Mistaking it for Rolle's when endpoints differ is another common issue. Choice C accurately captures this. A transferable strategy is to systematically check continuity on closed and differentiability on open intervals before applying the theorem.

Question 19

1. No; MVT requires $h$ to be constant on $[0,10]$.
2. Yes; because $h$ is continuous on $(0,10)$, there is a point where $h'(c)=0$.
3. Yes; $\frac{7-7}{10-0}=0$ and MVT hypotheses are met, so some $c$ has $h'(c)=0$. (correct answer)
4. No; differentiability at endpoints is required and not given.
5. Yes; endpoint equality alone guarantees a horizontal tangent without other conditions.

Explanation: When endpoints are equal, MVT implies a horizontal tangent via Rolle's Theorem. For h continuous on [0, 10] and differentiable on (0, 10), h(0) = 7 = h(10), the average is 0, guaranteeing h'(c) = 0. The theorem holds. A common misuse is thinking continuity on open suffices alone. Requiring differentiability at endpoints is another error. Choice C correctly identifies this. A transferable strategy is to use Rolle's as a MVT special case when secant slope is zero.

Question 20

1. Yes; because $g$ is continuous, it must have derivative $1$ somewhere.
2. No; MVT requires $g(-4)=g(4)$.
3. Yes; $\frac{9-1}{4-(-4)}=1$ and MVT hypotheses are satisfied. (correct answer)
4. No; MVT requires differentiability on $[-4,4]$.
5. Yes; Rolle’s Theorem guarantees $g'(c)=1$ since endpoints differ.

Explanation: The theorem applies, with average 1, guaranteeing c. Misuse: confusing with Rolle's for equal endpoints. Differentiability on closed not required. Choice C correct. Check average after conditions.