Mean Value Theorem
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AP Calculus AB › Mean Value Theorem
A function $f$ is continuous on $1,5$ with $f(1)=2$ and $f(5)=10$. Does the Mean Value Theorem guarantee a $c$ with $f'(c)=2$?
No, because knowing endpoint values alone never guarantees any derivative value.
Yes, because $f(1)$ and $f(5)$ are positive, so $f'(c)=2$ for some $c$.
Yes, because $f(5)-f(1)=8$, so there must be a point where $f'(c)=8$.
Yes, because $f$ is continuous on $[1,5]$, so $f'(c)$ must equal the average rate of change.
Yes, because $f$ is continuous on $[1,5]$ and differentiable on $(1,5)$, so some $c$ has $f'(c)=\dfrac{10-2}{5-1}=2$.
Explanation
The Mean Value Theorem applies when a function is continuous on a closed interval [a,b] and differentiable on the open interval (a,b). Since f satisfies both conditions on [1,5], MVT guarantees there exists some c in (1,5) where f'(c) equals the average rate of change. The average rate of change is (f(5)-f(1))/(5-1) = (10-2)/(4) = 2. Therefore, MVT does guarantee a point c where f'(c) = 2. A common error is thinking that knowing only endpoint values is insufficient - but when continuity and differentiability are given, MVT provides this powerful conclusion. Always check both MVT conditions before applying the theorem.
A function $p$ is continuous on $-1,2$ with $p(-1)=0$ and $p(2)=6$. Does the Mean Value Theorem guarantee a $c$ with $p'(c)=2$?
No, because MVT requires $p(-1)=p(2)$.
Yes, because $p(2)-p(-1)=6$.
Yes, because $p$ crosses 2.
No, because MVT requires $p$ to be increasing everywhere.
Yes, because $p$ is continuous on $[-1,2]$ and differentiable on $(-1,2)$, so some $c$ has $p'(c)=\dfrac{6-0}{2-(-1)}=2$.
Explanation
The Mean Value Theorem requires p to be continuous on [-1,2] and differentiable on (-1,2). Since these conditions are met, MVT guarantees there exists c in (-1,2) where p'(c) equals the average rate of change. The average rate of change is (p(2)-p(-1))/(2-(-1)) = (6-0)/3 = 6/3 = 2. Therefore, MVT does guarantee a point where p'(c) = 2. Students might think MVT requires functions to be increasing everywhere, but the theorem only requires the overall average behavior. The function can have various local behaviors while still satisfying MVT conditions.
A function $c$ is continuous on $-2,6$ with $c(-2)=3$ and $c(6)=11$. Does the Mean Value Theorem guarantee a $d$ with $c'(d)=1$?
Yes, because $c(6)-c(-2)=8$.
No, because MVT requires $c(-2)=c(6)$.
Yes, because $c$ is continuous, so $c'(d)=1$ at the midpoint.
Yes, because $c$ is continuous on $[-2,6]$ and differentiable on $(-2,6)$, so some $d$ has $c'(d)=\dfrac{11-3}{6-(-2)}=1$.
No, because MVT requires the interval to be $[0,6]$.
Explanation
The Mean Value Theorem requires c to be continuous on [-2,6] and differentiable on (-2,6). Since these conditions are satisfied, MVT guarantees there exists d in (-2,6) where c'(d) equals the average rate of change. Computing: (c(6)-c(-2))/(6-(-2)) = (11-3)/8 = 8/8 = 1. Therefore, MVT does guarantee a point d where c'(d) = 1. Students might think MVT requires specific interval forms like [0,6], but the theorem applies to any closed interval where the function meets the required conditions. The specific endpoints don't affect MVT's validity.
Let $t$ be continuous on $-3,1$ and differentiable on $(-3,1)$ with $t(-3)=0$ and $t(1)=8$. Does MVT guarantee a $c$ with $t'(c)=2$?
Yes, because $t$ is continuous on $[-3,1]$, differentiable on $(-3,1)$, and $\frac{t(1)-t(-3)}{1-(-3)}=\frac{8-0}{4}=2$.
No, because $t(-3)=0$, so the derivative must be $0$ somewhere, not $2$.
No, because MVT requires $t$ to be linear.
Yes, because $t$ is continuous on $[-3,1]$ and $2$ is between $t(-3)$ and $t(1)$.
Yes, because $t$ is defined on $[-3,1]$ and has an average slope of $2$.
Explanation
The function t satisfies MVT's requirements: continuous on [-3,1] and differentiable on (-3,1). Computing the average rate of change: (t(1)-t(-3))/(1-(-3)) = (8-0)/4 = 8/4 = 2. Since this average rate equals the desired derivative value of 2, MVT guarantees there exists at least one c in (-3,1) where t'(c) = 2. The fact that t(-3) = 0 doesn't mean the derivative must be 0 somewhere; that would only be true if both endpoints had the same function value (Rolle's Theorem). MVT doesn't require linearity; it works for any continuous, differentiable function. To apply MVT correctly, verify the two conditions and check if the average rate matches the target derivative.
A function $h$ is continuous on $2,6$ and differentiable on $(2,6)$ with $h(2)=-3$ and $h(6)=5$. Does MVT guarantee $c$ with $h'(c)=2$?
Yes, because $h$ is differentiable on $(2,6)$, so it must take every derivative value.
Yes, because $h$ is continuous on $[2,6]$ and $2$ is between $h(2)$ and $h(6)$.
No, because $h(2)$ is negative while $h(6)$ is positive.
Yes, because $h$ is continuous on $[2,6]$, differentiable on $(2,6)$, and $\frac{h(6)-h(2)}{6-2}=\frac{5-(-3)}{4}=2$.
No, because MVT requires $h(2)=h(6)$.
Explanation
To apply the Mean Value Theorem, we need continuity on [2,6] and differentiability on (2,6), both of which are given. The average rate of change over this interval is (h(6)-h(2))/(6-2) = (5-(-3))/4 = 8/4 = 2. Since this average rate equals the desired derivative value of 2, MVT guarantees there exists at least one c in (2,6) where h'(c) = 2. The fact that h changes from negative to positive values is irrelevant to MVT's application. A common error is confusing MVT with the Intermediate Value Theorem, which deals with function values rather than derivatives. Remember: MVT connects the average rate of change to an instantaneous rate of change somewhere in the interval.
A function $u$ is continuous on $1,9$ and differentiable on $(1,9)$ with $u(1)=-2$ and $u(9)=6$. Does MVT guarantee some $c$ where $u'(c)=1$?
No, because MVT requires $u(1)=u(9)$.
Yes, because $u$ is continuous on $[1,9]$ and $1$ is between $u(1)$ and $u(9)$.
Yes, because $u$ is differentiable on $(1,9)$, so $u'(c)$ equals $1$ for some $c$.
Yes, because $u$ is continuous on $[1,9]$, differentiable on $(1,9)$, and $\frac{u(9)-u(1)}{9-1}=\frac{6-(-2)}{8}=1$.
No, because $u(1)<u(9)$, so $u'(c)$ must be greater than $1$.
Explanation
The function u meets both MVT requirements: continuous on [1,9] and differentiable on (1,9). The average rate of change is (u(9)-u(1))/(9-1) = (6-(-2))/8 = 8/8 = 1, which matches exactly the derivative value we're looking for. By MVT, there must exist at least one c in (1,9) where u'(c) = 1. The fact that u(1) < u(9) means u increases overall with an average rate of 1, but doesn't mean u'(c) must always exceed 1. MVT doesn't require equal endpoint values; that's a special case (Rolle's Theorem). When applying MVT, calculate the average rate of change and verify it equals the desired derivative value to confirm MVT guarantees such a point exists.
Suppose $p$ is continuous on $1,5$ and differentiable on $(1,5)$ with $p(1)=7$ and $p(5)=3$. Does MVT guarantee a $c$ where $p'(c)=-1$?
Yes, because $p$ is defined on $[1,5]$, so the average rate of change must occur.
Yes, because $p$ is continuous on $[1,5]$ and $-1$ lies between $p(1)$ and $p(5)$.
Yes, because $p$ is continuous on $[1,5]$, differentiable on $(1,5)$, and $\frac{p(5)-p(1)}{5-1}=\frac{3-7}{4}=-1$.
No, because differentiability on $(1,5)$ is not enough to apply MVT.
No, because $p$ decreases, so $p'(x)$ cannot equal $-1$.
Explanation
The function p satisfies both MVT requirements: it's continuous on [1,5] and differentiable on (1,5). Computing the average rate of change: (p(5)-p(1))/(5-1) = (3-7)/4 = -4/4 = -1. This matches exactly the derivative value we're seeking. Therefore, MVT guarantees the existence of at least one c in (1,5) where p'(c) = -1. The fact that p is decreasing (since p(5) < p(1)) actually supports having a negative derivative, contrary to what choice C suggests. A common mistake is thinking that a decreasing function must have a constant derivative, but derivatives can vary while remaining negative. To apply MVT effectively, focus on verifying the two conditions and calculating the average rate of change.
Let $g$ be continuous on $-1,3$ and differentiable on $(-1,3)$ with $g(-1)=5$ and $g(3)=1$. Does MVT guarantee $c$ where $g'(c)=-1$?
Yes, because $g$ is differentiable somewhere on $(-1,3)$, so $g'(c)=-1$ for some $c$.
No, because $g$ is decreasing, so $g'(c)$ must be constant.
Yes, because $g$ is continuous on $[-1,3]$, differentiable on $(-1,3)$, and $\frac{g(3)-g(-1)}{3-(-1)}=\frac{1-5}{4}=-1$.
No, because $g(-1)\ne g(3)$, so MVT cannot be applied.
Yes, because $g$ is continuous on $[-1,3]$ and $-1$ lies between $g(-1)$ and $g(3)$.
Explanation
The Mean Value Theorem requires two conditions: continuity on the closed interval [a,b] and differentiability on the open interval (a,b). Here, both conditions are explicitly satisfied for g on [-1,3] and (-1,3) respectively. The average rate of change is (g(3)-g(-1))/(3-(-1)) = (1-5)/4 = -4/4 = -1, which is exactly what we're looking for. Since all MVT conditions are met and the average rate equals -1, MVT guarantees there exists at least one c in (-1,3) where g'(c) = -1. A common misconception is that MVT requires equal function values at endpoints (that's Rolle's Theorem), but MVT works for any continuous, differentiable function. When applying MVT, calculate the average rate of change first, then check if it matches the desired derivative value.
Suppose $s$ is continuous on $0,8$ and differentiable on $(0,8)$ with $s(0)=9$ and $s(8)=1$. Does MVT guarantee some $c$ with $s'(c)=-1$?
No, because $s(0)>s(8)$ implies $s'(x)$ is always negative but not necessarily $-1$.
No, because MVT does not apply on closed intervals.
Yes, because $s$ is differentiable on $(0,8)$, so it must match the average slope.
Yes, because $-1$ is between $s(0)$ and $s(8)$.
Yes, because $s$ is continuous on $[0,8]$, differentiable on $(0,8)$, and $\frac{s(8)-s(0)}{8-0}=\frac{1-9}{8}=-1$.
Explanation
The function s meets both MVT conditions: continuous on [0,8] and differentiable on (0,8). The average rate of change is (s(8)-s(0))/(8-0) = (1-9)/8 = -8/8 = -1, which is exactly the derivative value in question. Therefore, MVT guarantees the existence of at least one c in (0,8) where s'(c) = -1. While s is decreasing overall (since s(8) < s(0)), this doesn't mean s'(x) is constantly -1; the derivative could vary while remaining negative. A common mistake is thinking MVT doesn't apply to closed intervals, but it specifically requires continuity on the closed interval [a,b]. Remember: MVT connects average and instantaneous rates of change, regardless of whether the function increases or decreases.
A function $r$ is continuous on $3,7$ and differentiable on $(3,7)$ with $r(3)=4$ and $r(7)=12$. Does MVT guarantee a $c$ such that $r'(c)=2$?
Yes, because $2$ lies between $r(3)$ and $r(7)$.
Yes, because $r$ is continuous on $[3,7]$ (differentiability is unnecessary).
Yes, because $r$ is continuous on $[3,7]$, differentiable on $(3,7)$, and $\frac{r(7)-r(3)}{7-3}=\frac{12-4}{4}=2$.
No, because $r(7)$ is larger than $r(3)$, so $r'(c)$ must exceed $2$.
No, because MVT requires $r(3)=r(7)$.
Explanation
The function r satisfies MVT's requirements: continuous on [3,7] and differentiable on (3,7). Calculating the average rate of change: (r(7)-r(3))/(7-3) = (12-4)/4 = 8/4 = 2. This average rate matches exactly the derivative value we seek. By MVT, there must exist at least one c in (3,7) where r'(c) = 2. Choice C incorrectly applies the Intermediate Value Theorem logic to derivatives rather than function values. MVT doesn't require the derivative value to lie between endpoint function values; it guarantees the derivative equals the average rate of change. When solving MVT problems, focus on the average rate calculation rather than the relationship between function values.