Mean Value Theorem - AP Calculus BC

1. Let $g$ be continuous on $[-1,1]$, differentiable on $(-1,1)$, with $g(-1)=2$ and $g(1)=2$. Does MVT guarantee some $c$ where $g'(c)=0$?
   Answer: Yes; $\frac{2-2}{1-(-1)}=0$ and the hypotheses match MVT, so some $c$ has $g'(c)=0$. With symmetric endpoints and equal values, MVT via Rolle's ensures a zero derivative point. For $g$ continuous on $[-1, 1]$ and differentiable on $(-1, 1)$, $g(-1) = 2 = g(1)$, the rate is 0, guaranteeing $g'(c) = 0$. Conditions hold. A common misuse is requiring differentiability on closed interval. Thinking continuity alone suffices is another error. Choice C properly states this. A transferable strategy is to treat symmetric intervals with equal endpoints as prime for Rolle's application.
2. Suppose $h$ is continuous on $[2,12]$, differentiable on $(2,12)$, with $h(2)=20$ and $h(12)=0$. Does MVT guarantee a $c$ with $h'(c)=-2$?
   Answer: Yes; $\frac{0-20}{12-2}=-2$ and the hypotheses satisfy MVT. MVT states there's a c with f'(c) = secant slope if continuous on closed and differentiable on open interval. h meets this, with (0 - 20) / (12 - 2) = -2, so yes. Common error: assuming MVT needs differentiability at endpoints, but open interval is enough. Mixing with Rolle's by requiring differing endpoints for non-zero slope is misguided. Choice B is correct. Strategy: Confirm open differentiability separately from closed continuity for proper application.
3. Assume $t$ is continuous on $[-4,-1]$, differentiable on $(-4,-1)$, and $t(-4)=5$, $t(-1)=2$. Does MVT guarantee $t'(c)=-1$?
   Answer: Yes, because $t$ is continuous on $[-4,-1]$ and differentiable on $(-4,-1)$. 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.
4. Suppose $h$ is continuous on $[3,9]$, differentiable on $(3,9)$, with $h(3)=2$ and $h(9)=14$. Does MVT guarantee a $c$ with $h'(c)=2$?
   Answer: Yes; $\frac{14-2}{9-3}=2$ and MVT hypotheses are met. MVT holds, average 2, yes. Misuse: thinking concavity required, no. Equal endpoints not needed. Choice A accurate. Verify without extra assumptions.
5. Let $h$ be continuous on $[-3,1]$, differentiable on $(-3,1)$, with $h(-3)=-2$ and $h(1)=6$. Does MVT guarantee $c$ where $h'(c)=2$?
   Answer: Yes; $\frac{6-(-2)}{1-(-3)}=2$ and hypotheses match MVT. 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.
6. Suppose $h$ is continuous on $[-6,-2]$, differentiable on $(-6,-2)$, with $h(-6)=4$ and $h(-2)=0$. Does MVT guarantee a $c$ with $h'(c)=-1$?
   Answer: Yes; since $\frac{0-4}{-2-(-6)}=-1$ and MVT conditions are met. MVT operates on entirely negative intervals without issue. For h continuous on [-6, -2] and differentiable on (-6, -2), h(-6) = 4 and h(-2) = 0, the rate is (0 - 4) / (-2 - (-6)) = -1, guaranteeing h'(c) = -1. Conditions are satisfied. A common misuse is assuming positive intervals are needed. Requiring differentiability on closed is another mistake. Choice B properly applies MVT. A transferable strategy is to ensure correct subtraction order in negative intervals for accurate slopes.
7. Let $s$ be continuous on $[0,6]$ and differentiable on $(0,6)$ with $s(0)=-3$ and $s(6)=-3$. Does MVT guarantee a $c$ with $s'(c)=0$?
   Answer: Yes; because $s(0)=s(6)$, MVT guarantees some $c$ with $s'(c)=0$. The function s is continuous on [0,6] and differentiable on (0,6), meeting MVT's conditions. Since s(0) = s(6) = -3, the average rate of change is (s(6)-s(0))/(6-0) = (-3-(-3))/6 = 0/6 = 0. Therefore, MVT guarantees at least one c in (0,6) where s'(c) = 0. This means there's at least one horizontal tangent line in the interval. A common mistake is thinking equal endpoint values prevent MVT from applying, when actually this special case guarantees the existence of a critical point. When f(a) = f(b), MVT ensures at least one point where the derivative equals zero.
8. Let $p$ be continuous on $[6,10]$, differentiable on $(6,10)$, with $p(6)=14$ and $p(10)=6$. Does MVT guarantee a $c$ with $p'(c)=-2$?
   Answer: Yes; $\frac{6-14}{10-6}=-2$ and hypotheses satisfy MVT. MVT guarantees the existence of c for the average slope in qualifying scenarios. For p continuous on [6, 10] and differentiable on (6, 10), $p(6) = 14$ and $p(10) = 6$, the rate is $\frac{6-14}{10-6} = -2$, so $p'(c) = -2$ exists. Conditions are met. A common misuse is thinking equal endpoints are required. Another error is assuming continuity on open alone works. Choice B correctly applies this. A transferable strategy is to consistently verify both continuity and differentiability as specified before concluding.
9. Let $p$ be continuous on $[2,8]$, differentiable on $(2,8)$, with $p(2)=7$ and $p(8)=7$. Does MVT guarantee some $c$ where $p'(c)=0$?
   Answer: Yes; since $p(2)=p(8)$ and $p$ is continuous on $[2,8]$ and differentiable on $(2,8)$, some $c$ has $p'(c)=0$. When endpoint values are equal, MVT specializes to Rolle's Theorem, guaranteeing a point where the derivative is zero if the function is continuous on [a, b] and differentiable on (a, b). Here, p is continuous on [2, 8] and differentiable on (2, 8), with p(2) = 7 = p(8), so there exists c with p'(c) = 0. The application is valid as hypotheses are met. A common misuse is requiring the function to be increasing or differentiable at endpoints, neither of which is necessary. Confusing MVT with the Intermediate Value Theorem, which deals with function values, is another error. Choice A correctly applies this special case. A transferable strategy is to recognize when equal endpoints imply Rolle's and verify conditions accordingly for zero derivative guarantees.
10. Let $p$ be continuous on $[-2,6]$, differentiable on $(-2,6)$, with $p(-2)=9$ and $p(6)=1$. Does MVT guarantee $c$ where $p'(c)=-1$?
    Answer: Yes; $\frac{1-9}{6-(-2)}=-1$ and hypotheses satisfy MVT. 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.
11. Suppose $g$ is continuous on $[0,3]$, differentiable on $(0,3)$, $g(0)=9$, and $g(3)=0$. Does MVT guarantee a $c$ with $g'(c)=-3$?
    Answer: Yes; MVT guarantees $g'(c)=\frac{0-9}{3-0}=-3$ for some $c$. 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.
12. Suppose $f$ is continuous on $[-1,3]$, differentiable on $(-1,3)$, and $f(-1)=4$, $f(3)=-4$. Does MVT guarantee a $c$ with $f'(c)=-2$?
    Answer: Yes; the average slope is $\frac{-4-4}{3-(-1)}=-2$, and conditions match MVT. 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.
13. If $p$ is continuous on $[-1,2]$, differentiable on $(-1,2)$, with $p(-1)=3$ and $p(2)=0$, does MVT guarantee $p'(c)=-1$?
    Answer: Yes, because $p$ is continuous on $[-1,2]$ and differentiable on $(-1,2)$. The function p meets MVT's criteria: continuous on [-1,2] and differentiable on (-1,2). The average rate of change is (p(2)-p(-1))/(2-(-1)) = (0-3)/3 = -3/3 = -1. Thus, MVT guarantees there exists at least one c in (-1,2) where p'(c) = -1. Students sometimes think MVT requires the function to be monotonic (always decreasing or increasing), but this is false—the function can increase and decrease within the interval. The different signs of p(-1) and p(2) don't affect MVT's application. Key insight: MVT only requires the basic continuity and differentiability conditions, not specific behavior patterns.
14. Let $p$ be continuous on $[-1,7]$, differentiable on $(-1,7)$, with $p(-1)=3$ and $p(7)=19$. Does MVT guarantee $c$ where $p'(c)=2$?
    Answer: Yes; $\frac{19-3}{7-(-1)}=2$ and MVT hypotheses are met. 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.
15. A function $v$ is continuous on $[-1,1]$ and differentiable on $(-1,1)$ with $v(-1)=3$ and $v(1)=7$. Does MVT guarantee some $c$ with $v'(c)=2$?
    Answer: Yes; since $v$ is continuous on $[-1,1]$ and differentiable on $(-1,1)$, some $c$ has $v'(c)=\dfrac{7-3}{1-(-1)}=2$. The function v satisfies MVT's conditions: continuous on [-1,1] and differentiable on (-1,1). The average rate of change is (v(1)-v(-1))/(1-(-1)) = (7-3)/2 = 4/2 = 2. By MVT, there exists at least one c in (-1,1) where v'(c) = 2. A common error is thinking MVT requires equal endpoint values or differentiability on the closed interval. The theorem only needs continuity on [a,b] and differentiability on (a,b). Remember the strategy: verify the two conditions, calculate the average rate of change, and that value is what f'(c) must equal at some point in the open interval.
16. Suppose $h$ is continuous on $[0,10]$, differentiable on $(0,10)$, with $h(0)=7$ and $h(10)=7$. Does MVT guarantee a $c$ with $h'(c)=0$?
    Answer: Yes; $\frac{7-7}{10-0}=0$ and MVT hypotheses are met, so some $c$ has $h'(c)=0$. 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.
17. If $f$ is continuous on $[7,9]$, differentiable on $(7,9)$, with $f(7)=3$ and $f(9)=11$, does MVT guarantee a $c$ with $f'(c)=4$?
    Answer: Yes; $\frac{11-3}{9-7}=4$ and MVT hypotheses are satisfied. 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.
18. Given $g$ continuous on $[-2,2]$, differentiable on $(-2,2)$, $g(-2)=5$, and $g(2)=1$, does MVT guarantee $c$ with $g'(c)=-1$?
    Answer: Yes; continuity on $[-2,2]$ and differentiability on $(-2,2)$ guarantee some $c$ with $g'(c)=\frac{1-5}{2-(-2)}=-1$. The Mean Value Theorem guarantees that for a function continuous on [a, b] and differentiable on (a, b), there is a c where the derivative equals the secant slope over the interval. Here, g is continuous on [-2, 2] and differentiable on (-2, 2), with g(-2) = 5 and g(2) = 1, yielding an average rate of (1 - 5) / (2 - (-2)) = -1, so MVT ensures a c with g'(c) = -1. The conditions are precisely those required, making the conclusion valid. A frequent misuse is thinking MVT needs differentiability at endpoints, which it does not. People often mistakenly invoke Rolle's Theorem when endpoints differ, but here MVT applies directly. This explains why choice C correctly identifies the application. A transferable strategy is to calculate the average rate first, then confirm the hypotheses to ensure MVT's guarantee holds.
19. Let $g$ be continuous on $[1,9]$, differentiable on $(1,9)$, with $g(1)=6$ and $g(9)=-2$. Does MVT guarantee $c$ where $g'(c)=-1$?
    Answer: Yes; $\frac{-2-6}{9-1}=-1$ and the hypotheses match MVT. For functions with differing signs at endpoints, MVT still applies if conditions hold. For g continuous on [1, 9] and differentiable on (1, 9), g(1) = 6 and g(9) = -2, the average is (-2 - 6) / (9 - 1) = -1, guaranteeing g'(c) = -1. The theorem is valid here. A common misuse is thinking MVT requires positive values. Another error is misapplying Rolle's when endpoints differ. Choice B correctly identifies this. A transferable strategy is to disregard function sign and focus on MVT's core requirements.
20. If $f$ is continuous on $[1,5]$, differentiable on $(1,5)$, and $f(1)=2$, $f(5)=10$, does MVT guarantee $f'(c)=2$?
    Answer: Yes, because $f$ is continuous on $[1,5]$ and differentiable on $(1,5)$. The Mean Value Theorem (MVT) states that if a function is continuous on [a,b] and differentiable on (a,b), then there exists at least one c in (a,b) where f'(c) equals the average rate of change (f(b)-f(a))/(b-a). Here, f is continuous on [1,5] and differentiable on (1,5), so MVT applies. The average rate of change is (10-2)/(5-1) = 8/4 = 2, so MVT guarantees there exists some c where f'(c) = 2. A common mistake is thinking MVT requires equal function values at endpoints (that's Rolle's Theorem) or that the function must have specific properties like being positive. The key strategy: check continuity on the closed interval and differentiability on the open interval—if both hold, MVT guarantees the existence of c.

Start this quiz