Mean Value Theorem
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AP Calculus BC › Mean Value Theorem
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 $t'(c)=-1$?
No; MVT requires $t$ to be differentiable at $x=-4$ and $x=2$.
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; because $t(-4)>t(2)$, there must be $c$ with $t'(c)=-1$ even without continuity.
Yes; continuity on $(-4,2)$ alone guarantees a point where $t'(c)=-1$.
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 $f'(c)=4$?
Yes; $\frac{11-3}{9-7}=4$ and MVT hypotheses are satisfied.
Yes; endpoint values guarantee a point with derivative equal to the larger endpoint value.
No; MVT applies only when $f(7)=f(9)$.
No; MVT requires $f$ to be differentiable on $[7,9]$.
Yes; because $f$ is continuous on $(7,9)$, it must have derivative $4$.
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; because $p$ is continuous, it must have a point with slope $-2$.
Yes; $\frac{1-5}{2-0}=-2$ and the hypotheses satisfy MVT, so such $c$ exists.
No; MVT requires $p(0)=p(2)$.
Yes; differentiability on $[0,2]$ alone guarantees it.
No; MVT requires $p$ to be decreasing on the entire interval.
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.
Let $g$ be continuous on $1,6$, differentiable on $(1,6)$, with $g(1)=0$ and $g(6)=15$. Does MVT guarantee some $c$ where $g'(c)=3$?
Yes; continuity on $[1,6]$ alone guarantees a point where $g'(c)=3$.
Yes; since $\frac{15-0}{6-1}=3$ and MVT hypotheses hold.
Yes; differentiability on $(1,6)$ alone guarantees such a $c$.
No; MVT requires $g(1)=g(6)$.
No; MVT requires $g$ to be increasing everywhere.
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.
Suppose $h$ is continuous on $1,4$, differentiable on $(1,4)$, with $h(1)=-1$ and $h(4)=8$. Does MVT guarantee $c$ where $h'(c)=3$?
Yes; because $h$ is differentiable on $(1,4)$, it must take slope $3$.
No; continuity on $(1,4)$ is required instead of $[1,4]$.
Yes; $\frac{8-(-1)}{4-1}=3$ and MVT conditions are satisfied.
Yes; any continuous function has a point where derivative equals $3$.
No; MVT requires $h(1)=h(4)$.
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.
Let $p$ be continuous on $-2,6$, differentiable on $(-2,6)$, with $p(-2)=9$ and $p(6)=1$. Does MVT guarantee $c$ where $p'(c)=-1$?
No; MVT requires $p(-2)=p(6)$.
No; MVT cannot be applied when the interval crosses $0$.
Yes; $\frac{1-9}{6-(-2)}=-1$ and hypotheses satisfy MVT.
Yes; because $p$ is continuous, a slope of $-1$ must occur.
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.
Let $p$ be continuous on $-1,7$, differentiable on $(-1,7)$, with $p(-1)=3$ and $p(7)=19$. Does MVT guarantee $c$ where $p'(c)=2$?
Yes; differentiability on $(-1,7)$ alone guarantees it.
No; MVT requires differentiability on $[-1,7]$.
Yes; $\frac{19-3}{7-(-1)}=2$ and MVT hypotheses are met.
Yes; because $p$ is continuous, it must have derivative $2$ somewhere.
No; MVT requires $p(-1)=p(7)$.
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.
Suppose $f$ is continuous on $-1,3$, differentiable on $(-1,3)$, and $f(-1)=4$, $f(3)=-4$. Does MVT guarantee a $c$ with $f'(c)=-2$?
Yes; the average slope is $\frac{-4-4}{3-(-1)}=-2$, and conditions match MVT.
No; MVT requires $f(-1)=f(3)$.
Yes; because $f$ is differentiable on $(-1,3)$, some $c$ has $f'(c)=-2$.
No; MVT requires $f$ to be continuous on $(-1,3)$ only.
No; MVT cannot be used when function values are negative.
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.
Suppose $g$ is continuous on $0,3$, differentiable on $(0,3)$, $g(0)=9$, and $g(3)=0$. Does MVT guarantee a $c$ with $g'(c)=-3$?
No; MVT applies only if $g$ is linear.
Yes; because $g$ is continuous, it must have derivative $-3$ somewhere.
Yes; MVT guarantees $g'(c)=\frac{0-9}{3-0}=-3$ for some $c$.
Yes; Rolle’s Theorem applies since $g(0)\neq g(3)$.
No; MVT requires $g$ to be differentiable at $0$ and $3$.
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.
If $f$ is continuous on $0,5$, differentiable on $(0,5)$, with $f(0)=12$ and $f(5)=2$, does MVT guarantee a $c$ with $f'(c)=-2$?
No; MVT requires $f$ to be decreasing everywhere.
Yes; $\frac{2-12}{5-0}=-2$ and the hypotheses satisfy MVT.
Yes; because $f$ is differentiable, it must take slope $-2$.
No; MVT applies only when $f(0)=f(5)$.
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.