Mean Value Theorem
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AP Calculus AB › Mean Value Theorem
A function $t$ is continuous on $0,1$ with $t(0)=0$ and $t(1)=5$. Does the Mean Value Theorem guarantee a $c$ with $t'(c)=5$?
Yes, because $t$ is continuous on $[0,1]$ and differentiable on $(0,1)$, so some $c$ has $t'(c)=\dfrac{5-0}{1-0}=5$.
No, because MVT requires $t(0)=t(1)$.
Yes, because $t(1)=5$.
No, because MVT does not apply on intervals of length 1.
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.
A function $f$ is continuous on $1,13$ with $f(1)=5$ and $f(13)=17$. Does the Mean Value Theorem guarantee a $c$ with $f'(c)=1$?
Yes, because $f$ is continuous so $f'(c)=1$ at the midpoint.
Yes, because $f(13)-f(1)=12$.
Yes, because $f$ is continuous on $[1,13]$ and differentiable on $(1,13)$, so some $c$ has $f'(c)=\dfrac{17-5}{13-1}=1$.
No, because MVT requires $f(1)=f(13)$.
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.
A function $g$ is continuous on $-2,4$ with $g(-2)=7$ and $g(4)=1$. Does the Mean Value Theorem guarantee a $c$ with $g'(c)=-1$?
No, because the Mean Value Theorem requires $g$ to be increasing.
No, because $g(-2)\neq g(4)$, so the Mean Value Theorem cannot apply.
Yes, because $g$ changes from 7 to 1, so $g'(c)=-1$ somewhere.
Yes, because $g$ is continuous on $[-2,4]$ and differentiable on $(-2,4)$, so some $c$ has $g'(c)=\dfrac{1-7}{4-(-2)}=-1$.
Yes, because $g$ is differentiable on $[-2,4]$, so $g'(c)$ equals $g(4)-g(-2)$.
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.
A function $f$ is continuous on $2,8$ with $f(2)=5$ and $f(8)=5$. Does the Mean Value Theorem guarantee a $c$ with $f'(c)=0$?
No, because MVT requires $f(2)\ne f(8)$.
Yes, because equal endpoint values guarantee $f$ has a minimum at the midpoint.
No, because $f$ might be constant only at the endpoints.
Yes, because $f$ is continuous, so $f'(c)=0$ must occur.
Yes, because $f(2)=f(8)$, so the average rate of change is 0 and MVT gives $f'(c)=0$ for some $c$.
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.
A function $G$ is continuous on $2,6$ with $G(2)=-1$ and $G(6)=7$. Does the Mean Value Theorem guarantee a $c$ with $G'(c)=2$?
No, because MVT requires $G(2)=G(6)$.
No, because $G(2)$ is negative.
Yes, because $G$ crosses 0.
Yes, because $G$ is continuous, so the derivative must be constant.
Yes, because $G$ is continuous on $[2,6]$ and differentiable on $(2,6)$, so some $c$ has $G'(c)=\dfrac{7-(-1)}{6-2}=2$.
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.
A function $S$ is continuous on $-6,-2$ with $S(-6)=-4$ and $S(-2)=4$. Does the Mean Value Theorem guarantee a $c$ with $S'(c)=2$?
Yes, because $S$ changes sign.
Yes, because $S$ is continuous so its derivative must be 2 at the midpoint.
No, because MVT requires $S(-6)=S(-2)$.
No, because negative inputs prevent using MVT.
Yes, because $S$ is continuous on $[-6,-2]$ and differentiable on $(-6,-2)$, so some $c$ has $S'(c)=\dfrac{4-(-4)}{-2-(-6)}=2$.
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.
A function $R$ is continuous on $0,8$ with $R(0)=2$ and $R(8)=18$. Does the Mean Value Theorem guarantee a $c$ with $R'(c)=2$?
No, because the derivative might not exist at endpoints.
Yes, because $R$ is continuous on $[0,8]$ and differentiable on $(0,8)$, so some $c$ has $R'(c)=\dfrac{18-2}{8-0}=2$.
Yes, because $R(8)-R(0)=16$.
No, because MVT only works if $R$ is decreasing.
Yes, because $R$ is continuous, so $R'(c)=2$ for all $c$.
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.
A function $q$ is continuous on $4,6$ with $q(4)=1$ and $q(6)=9$. Does the Mean Value Theorem guarantee a $c$ with $q'(c)=4$?
Yes, because $q$ is continuous on $[4,6]$ and differentiable on $(4,6)$, so some $c$ has $q'(c)=\dfrac{9-1}{6-4}=4$.
Yes, because $q(6)-q(4)=8$.
No, because the derivative cannot exceed 1 on a short interval.
No, because MVT requires $q(4)=q(6)$.
Yes, because $q$ is continuous, so its derivative must be 4 at $x=5$.
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.
A function $n$ is continuous on $-4,0$ with $n(-4)=2$ and $n(0)=10$. Does the Mean Value Theorem guarantee a $c$ with $n'(c)=2$?
Yes, because $n$ is continuous on $[-4,0]$ and differentiable on $(-4,0)$, so some $c$ has $n'(c)=\dfrac{10-2}{0-(-4)}=2$.
Yes, because $n(0)-n(-4)=8$.
Yes, because $n$ is continuous, so it has slope 2 everywhere.
No, because the interval ends at 0.
No, because MVT requires $n(-4)=n(0)$.
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.
A function $T$ is continuous on $2,5$ with $T(2)=1$ and $T(5)=7$. Does the Mean Value Theorem guarantee a $c$ with $T'(c)=2$?
Yes, because $T$ is continuous, so $T'(c)=2$ must occur.
No, because MVT needs $T$ to be linear.
Yes, because $T$ is continuous on $[2,5]$ and differentiable on $(2,5)$, so some $c$ has $T'(c)=\dfrac{7-1}{5-2}=2$.
Yes, because $T(5)-T(2)=6$.
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.