Mean Value Theorem
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AP Calculus BC › Mean Value Theorem
Let $p$ be continuous on $-2,6$, differentiable on $(-2,6)$, with $p(-2)=9$ and $p(6)=1$. Does MVT guarantee $c$ where $p'(c)=-1$?
Yes; because $p$ is continuous, a slope of $-1$ must occur.
Yes; $\frac{1-9}{6-(-2)}=-1$ and hypotheses satisfy MVT.
No; MVT requires $p(-2)=p(6)$.
Yes; differentiability on $(-2,6)$ alone guarantees it.
No; MVT cannot be applied when the interval crosses $0$.
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.
Let $h$ be continuous on $-3,1$, differentiable on $(-3,1)$, with $h(-3)=-2$ and $h(1)=6$. Does MVT guarantee $c$ where $h'(c)=2$?
Yes; continuity on $[-3,1]$ alone guarantees it.
Yes; because $h$ is differentiable on $(-3,1)$, it must have slope $2$ somewhere.
No; MVT requires $h(-3)=h(1)$.
No; MVT cannot be applied when the left endpoint is negative.
Yes; $\frac{6-(-2)}{1-(-3)}=2$ and hypotheses match MVT.
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.
If $f$ is continuous on $1,2$, differentiable on $(1,2)$, with $f(1)=0$ and $f(2)=4$, does MVT guarantee $c$ where $f'(c)=4$?
Yes; because $f$ is continuous on $(1,2)$, the derivative must equal $4$ somewhere.
Yes; since $\frac{4-0}{2-1}=4$ and MVT conditions are satisfied, such $c$ exists.
No; the interval length is too small for MVT to apply.
Yes; any differentiable function has a point where $f'(c)$ equals $f(2)$.
No; MVT requires $f(1)=f(2)$.
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.
Let $q$ be continuous on $2,8$ and differentiable on $(2,8)$ with $q(2)=7$ and $q(8)=1$. Does MVT guarantee some $c$ with $q'(c)=-1$?
Yes; continuity on $[2,8]$ alone guarantees a point where $q'(c)=-1$.
No; MVT requires $q$ to be differentiable at $x=2$ and $x=8$.
Yes; since $q(2)>q(8)$, there must be $c$ with $q'(c)=-1$ even if $q$ is not differentiable.
Yes; because $q$ is continuous on $[2,8]$ and differentiable on $(2,8)$, some $c$ has $q'(c)=\dfrac{1-7}{8-2}=-1$.
No; MVT requires $q(2)=q(8)$.
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.
Suppose $q$ is continuous on $2,6$, differentiable on $(2,6)$, and $q(2)=9$, $q(6)=9$. Does MVT guarantee a $c$ with $q'(c)=0$?
Yes, because $q(2)=q(6)$ implies $q$ is constant on $[2,6]$.
Yes, because $q$ is differentiable at $x=2$ and $x=6$.
No, because $q$ might not have a maximum value on $[2,6]$.
No, because the Mean Value Theorem requires $q(2)\ne q(6)$.
Yes, because $q(2)=q(6)$ and $q$ is continuous on $[2,6]$ and differentiable on $(2,6)$.
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).
Assume $t$ is continuous on $-4,-1$, differentiable on $(-4,-1)$, and $t(-4)=5$, $t(-1)=2$. Does MVT guarantee $t'(c)=-1$?
Yes, because $t$ is continuous on $[-4,-1]$ and differentiable on $(-4,-1)$.
No, because the Mean Value Theorem requires $t$ to be differentiable at $x=-4$ and $x=-1$.
Yes, because $t(-4)$ and $t(-1)$ are both positive.
Yes, because $t$ has an average rate of change of $-1$ and is differentiable somewhere.
No, because $t$ must cross the $x$-axis on $[-4,-1]$.
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.
Let $u$ be continuous on $-4,0$ and differentiable on $(-4,0)$ with $u(-4)=-1$ and $u(0)=7$. Does MVT guarantee $c$ with $u'(c)=2$?
No; MVT cannot be used when an endpoint is $0$.
No; MVT requires $u(-4)=u(0)$.
Yes; because $u$ is differentiable on $[-4,0]$, MVT gives $u'(c)=2$.
Yes; $u$ is continuous on $[-4,0]$ and differentiable on $(-4,0)$, so some $c$ satisfies $u'(c)=\dfrac{7-(-1)}{0-(-4)}=2$.
Yes; continuity on $(-4,0)$ is sufficient to guarantee $u'(c)=2$.
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.
Suppose $t$ is continuous on $-4,2$ and differentiable on $(-4,2)$ with $t(-4)=10$ and $t(2)=4$. Does MVT guarantee some $c$ with $t'(c)=-1$?
Yes; because $t(-4)>t(2)$, there must be $c$ with $t'(c)=-1$ even without continuity.
Yes; since $t$ is continuous on $[-4,2]$ and differentiable on $(-4,2)$, some $c$ has $t'(c)=\dfrac{4-10}{2-(-4)}=-1$.
Yes; continuity on $(-4,2)$ alone guarantees a point where $t'(c)=-1$.
No; MVT requires $t$ to be differentiable at $x=-4$ and $x=2$.
No; MVT applies only when the function is linear.
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.
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 $f'(c)=4$?
No; MVT applies only when $f(7)=f(9)$.
Yes; endpoint values guarantee a point with derivative equal to the larger endpoint value.
Yes; $\frac{11-3}{9-7}=4$ and MVT hypotheses are satisfied.
Yes; because $f$ is continuous on $(7,9)$, it must have derivative $4$.
No; MVT requires $f$ to be differentiable on $[7,9]$.
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.
Let $p$ be continuous on $0,2$, differentiable on $(0,2)$, with $p(0)=5$ and $p(2)=1$. Does MVT guarantee $c$ where $p'(c)=-2$?
Yes; $\frac{1-5}{2-0}=-2$ and the hypotheses satisfy MVT, so such $c$ exists.
No; MVT requires $p$ to be decreasing on the entire interval.
Yes; differentiability on $[0,2]$ alone guarantees it.
No; MVT requires $p(0)=p(2)$.
Yes; because $p$ is continuous, it must have a point with slope $-2$.
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.