AP Calculus AB Quiz: Working With The Intermediate Value Theorem

What this quiz covers

This quiz focuses on Working With The Intermediate Value Theorem, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. There exists $c\in(-1,3)$ such that $R(c)=6$. (correct answer)
2. There exists $c\in(-1,3)$ such that $R(c)=12$.
3. There exists $c\in(-1,3)$ such that $R(c)=0$.
4. There exists $c\in(3,5)$ such that $R(c)=6$.
5. No value is guaranteed because $R(-1)$ and $R(3)$ are both positive.

Explanation: The Intermediate Value Theorem (IVT) asserts that a continuous function on [a,b] takes every value between f(a) and f(b) inside (a,b). R is continuous on [-1,3] with R(-1) = 2 and R(3) = 10, guaranteeing values between 2 and 10. 6 is between them, making choice A correct. Choice B is incorrect as 12 > 10, and choice C because 0 < 2. A common error is thinking same-sign endpoints prevent IVT, but it applies to any value in the range, refuting choice E. Wrong intervals invalidate choice D. Transferable IVT checklist: 1. Confirm continuity on [a,b]. 2. Verify k is between f(a) and f(b). 3. Ensure c is in (a,b).

Question 2

1. $h(c)=10$
2. $h(c)=-1$
3. $h(c)=5$ (correct answer)
4. $h(c)=0$
5. $h(c)=9$ only at $c=5$

Explanation: The Intermediate Value Theorem (IVT) applies to the continuous function h on [0, 5], where h(0) = 1 and h(5) = 9, guaranteeing every value between 1 and 9 is achieved. The value 5 is strictly between 1 and 9, so there exists c in [0, 5] with h(c) = 5. Since 5 ≠ 1 and 5 ≠ 9, c must be in (0, 5). This shows how IVT ensures intermediate values are hit in the interior when strictly between endpoints. A common error is selecting a value outside the range, like 10 or -1, which IVT does not guarantee. Confusing IVT with the Mean Value Theorem by requiring differentiability is another mistake. Transferable IVT checklist: 1. Confirm the function is continuous on [a, b]. 2. Verify k is between f(a) and f(b). 3. If yes, there exists c in [a, b] with f(c) = k; if k is strictly between and ≠ f(a), ≠ f(b), then c is in (a, b).

Question 3

1. There exists $c\in(1,4)$ such that $f(c)=0$. (correct answer)
2. There exists $c\in(1,4)$ such that $f(c)=5$.
3. There exists $c\in(0,1)$ such that $f(c)=0$.
4. There exists $c\in(1,4)$ such that $f(c)=-4$.
5. No such $c$ is guaranteed because $f$ might be discontinuous on $(1,4)$.

Explanation: The Intermediate Value Theorem (IVT) states that for a continuous function f on a closed interval [a, b], any value k between f(a) and f(b) is achieved at some c in [a, b]. Here, f(1) = -3 and f(4) = 2, with -3 < 0 < 2, so IVT guarantees a c in [1, 4] where f(c) = 0. Since 0 ≠ f(1) and 0 ≠ f(4), this c must be in the open interval (1, 4). This application shows that the function must cross the x-axis between 1 and 4 due to the sign change. A common error is assuming IVT applies without a sign change or without continuity, but here both conditions are met. Another mistake is thinking IVT guarantees exactly one root, whereas it only ensures at least one. Transferable IVT checklist: 1. Confirm the function is continuous on [a, b]. 2. Verify k is between f(a) and f(b). 3. If yes, there exists c in [a, b] with f(c) = k; if k is strictly between and ≠ f(a), ≠ f(b), then c is in (a, b).

Question 4

1. There exists $c\in(-1,1)$ such that $A(c)=4$. (correct answer)
2. There exists $c\in(-1,1)$ such that $A(c)=0$.
3. There exists $c\in(-1,1)$ such that $A(c)=7$.
4. There exists $c\in(1,3)$ such that $A(c)=4$.
5. No value is guaranteed because the interval is symmetric.

Explanation: The Intermediate Value Theorem (IVT) ensures $A$ continuous on $[-1,1]$ with $A(-1) = 2$ and $A(1) = 6$ takes all values between 2 and 6 in $(-1,1)$. 4 is between them, so choice A is correct. Choice C fails as 7 > 6, and choice B because 0 < 2. A common error is assuming symmetry affects IVT, but it doesn't, refuting choice E. Wrong intervals mislead in choice D. Transferable IVT checklist: 1. Confirm continuity on $[a,b]$. 2. Verify k is between $f(a)$ and $f(b)$. 3. Ensure c is in $(a,b)$.

Question 5

1. There exists $c\in(-6,-2)$ such that $C(c)=-6$. (correct answer)
2. There exists $c\in(-6,-2)$ such that $C(c)=0$.
3. There exists $c\in(-6,-2)$ such that $C(c)=-12$.
4. There exists $c\in(-2,0)$ such that $C(c)=-6$.
5. No value is guaranteed since both endpoints are negative.

Explanation: The Intermediate Value Theorem (IVT) guarantees C continuous with C(-6) = -2 and C(-2) = -10 takes values between -10 and -2 in (-6,-2). -6 is between them, so choice A is correct. Choice C fails as -12 < -10, and choice B because 0 > -2. A common error is assuming same-sign endpoints prevent guarantees, but IVT applies to intermediates, refuting choice E. Wrong intervals invalidate choice D. Transferable IVT checklist: 1. Confirm continuity on [a,b]. 2. Verify k is between f(a) and f(b). 3. Ensure c is in (a,b).

Question 6

1. Some $c\in(-1,5)$ with $J(c)=-12$.
2. Some $c\in(-1,5)$ with $J(c)=0$. (correct answer)
3. Some $c\in(-1,5)$ with $J(c)=6$.
4. Some $c\in(5,7)$ with $J(c)=0$.
5. No conclusion can be drawn unless $J$ is polynomial.

Explanation: The Intermediate Value Theorem (IVT) ensures J continuous on [-1,5] with J(-1) = 3 and J(5) = -9 attains all values between -9 and 3 in (-1,5). 0 is between them, so choice B is correct. Choice C fails as 6 > 3, and choice A because -12 < -9. A common error is requiring a polynomial for IVT, but continuity suffices, refuting choice E. Wrong intervals mislead in choice D. Transferable IVT checklist: 1. Confirm continuity on [a,b]. 2. Verify k is between f(a) and f(b). 3. Ensure c is in (a,b).

Question 7

1. There exists $c\in(0,1)$ such that $f(c)=3$.
2. There exists $c\in(0,1)$ such that $f(c)=0$. (correct answer)
3. There exists $c\in(1,2)$ such that $f(c)=0$.
4. There exists $c\in(0,1)$ such that $f(c)=-2$.
5. No conclusion can be made because $f(0)\ne f(1)$.

Explanation: The Intermediate Value Theorem (IVT) ensures continuous f with f(0) = -1 and f(1) = 2 takes values between -1 and 2. Since -1 < 0 < 2, there exists c in [0, 1] with f(c) = 0, and as 0 ≠ -1, 0 ≠ 2, c is in (0, 1). This is a root guarantee from opposite signs. A common error is choosing 3, outside the range. Unequal endpoints do not prevent application; they enable it here. Transferable IVT checklist: 1. Confirm the function is continuous on [a, b]. 2. Verify k is between f(a) and f(b). 3. If yes, there exists c in [a, b] with f(c) = k; if k is strictly between and ≠ f(a), ≠ f(b), then c is in (a, b).

Question 8

1. There exists $c\in(1,5)$ such that $q(c)=-2$.
2. There exists $c\in(1,5)$ such that $q(c)=1$.
3. There exists $c\in(1,5)$ such that $q(c)=0$. (correct answer)
4. There exists $c\in(1,5)$ such that $q(c)=\tfrac{3}{2}$.
5. There exists $c\in(1,5)$ such that $q(c)=-3$.

Explanation: Given q continuous on [1,5] with q(1) = -3/2 and q(5) = 1/2, we need to find which value IVT guarantees. The function takes values from -3/2 to 1/2, and since these have opposite signs, 0 must be an intermediate value. This makes C the correct answer. Common errors include not recognizing that 0 always lies between a negative and positive value, or checking values like -2 or -3 which are outside the range [-3/2, 1/2]. The value 3/2 is also too large. IVT Checklist: ✓ Continuous function q, ✓ Endpoints have opposite signs, ✓ Zero is guaranteed between negative and positive values.

Question 9

1. There exists $c\in(2,8)$ such that $m(c)=\sqrt{2}$.
2. There exists $c\in(2,8)$ such that $m(c)=6$.
3. There exists $c\in(2,8)$ such that $m(c)=\tfrac{1}{2}$.
4. There exists $c\in(2,8)$ such that $m(c)=\sqrt{26}$.
5. There exists $c\in(2,8)$ such that $m(c)=3$. (correct answer)

Explanation: Given m continuous on [2,8] with m(2) = √2 ≈ 1.414 and m(8) = 5, IVT guarantees values between √2 and 5. Checking the options: 1/2 ≈ 0.5 is less than √2, while 6 exceeds 5, and √26 ≈ 5.099 is also greater than 5. Both √2 (option A) and 3 (option E) fall within the range, but option A is an endpoint value, not an intermediate value. Option E with value 3 is the only intermediate value in the range [√2, 5], making it correct. IVT Checklist: ✓ m is continuous, ✓ Range is [√2, 5] ≈ [1.414, 5], ✓ Value 3 is within this range.

Question 10

1. There exists $c\in(-1,2)$ such that $p(c)=-2$. (correct answer)
2. There exists $c\in(-1,2)$ such that $p(c)=0$.
3. There exists $c\in(-1,2)$ such that $p(c)=1$.
4. There exists $c\in(-1,2)$ such that $p(c)=-5$.
5. No value is guaranteed since $p(-1)$ and $p(2)$ are both negative.

Explanation: Here p is continuous on [-1,2] with p(-1) = -4 and p(2) = -1. Even though both endpoint values are negative, IVT still applies to any value between -4 and -1. Since -2 is in the range [-4,-1], IVT guarantees there exists c in (-1,2) where p(c) = -2, making A correct. A common misconception is thinking IVT requires endpoint values to have opposite signs (for finding zeros) - but IVT works for any intermediate value regardless of signs. Option E incorrectly suggests no value is guaranteed, while -5 is outside the range. IVT Checklist: ✓ Continuous function, ✓ Both endpoints negative is fine, ✓ Target -2 is between -4 and -1.

Question 11

1. A value $c\in(0,8)$ such that $y(c)=20$.
2. A value $c\in(0,8)$ such that $y(c)=4$. (correct answer)
3. A value $c\in(8,10)$ such that $y(c)=4$.
4. A value $c\in(0,8)$ such that $y(c)=-10$.
5. No conclusion can be drawn without the graph of $y$.

Explanation: The Intermediate Value Theorem (IVT) states that for continuous y on [0,8] with y(0) = -4 and y(8) = 12, every value between -4 and 12 is attained in (0,8). 4 is between them, so choice B is correct. Choice A fails because 20 > 12, and choice D as -10 < -4. A common error is demanding a graph, but IVT relies on continuity and endpoints alone, refuting choice E. Wrong intervals invalidate choice C. Another mistake is ignoring sign changes for non-root values. Transferable IVT checklist: 1. Confirm continuity on [a,b]. 2. Verify k is between f(a) and f(b). 3. Ensure c is in (a,b).

Question 12

1. There exists $c\in(-2,1)$ such that $f(c)=3$. (correct answer)
2. There exists $c\in(-2,1)$ such that $f(c)=0$.
3. There exists $c\in(-2,1)$ such that $f(c)=6$.
4. There exists $c\in(1,3)$ such that $f(c)=3$.
5. No value is guaranteed because $f$ might be discontinuous on $[-2,1]$.

Explanation: The Intermediate Value Theorem (IVT) guarantees that continuous functions on [a,b] hit every value between f(a) and f(b) in (a,b). f is continuous on [-2,1] with f(-2) = 5 and f(1) = 1, so values between 1 and 5 are assured. 3 is between them, making choice A correct. Choice C is wrong as 6 > 5, and choice B because 0 < 1. A common error is doubting continuity, but the problem states it, countering choice E. Wrong intervals appear in choice D. Transferable IVT checklist: 1. Confirm continuity on [a,b]. 2. Verify k is between f(a) and f(b). 3. Ensure c is in (a,b).

Question 13

1. A break-even point $c\in(1,3)$ with $P(c)=0$. (correct answer)
2. A point $c\in(1,3)$ with $P(c)=10$.
3. A point $c\in(0,1)$ with $P(c)=0$.
4. A point $c\in(1,3)$ with $P(c)=-6$.
5. No conclusion can be drawn because $P$ could jump between $x=1$ and $x=3$.

Explanation: The Intermediate Value Theorem (IVT) for continuous P on [1, 3] with P(1) = -5 and P(3) = 7 guarantees values between -5 and 7. Since -5 < 0 < 7, there exists c in [1, 3] with P(c) = 0, and as 0 ≠ -5, 0 ≠ 7, c is in (1, 3). This ensures a break-even point. A common error is thinking discontinuity allows jumping over values, but continuity prevents that. Selecting 10, outside the range, is incorrect. Transferable IVT checklist: 1. Confirm the function is continuous on [a, b]. 2. Verify k is between f(a) and f(b). 3. If yes, there exists c in [a, b] with f(c) = k; if k is strictly between and ≠ f(a), ≠ f(b), then c is in (a, b).

Question 14

1. A value $c\in(-4,-1)$ such that $W(c)=0$. (correct answer)
2. A value $c\in(-4,-1)$ such that $W(c)=7$.
3. A value $c\in(-4,-1)$ such that $W(c)=-3$.
4. A value $c\in(-1,2)$ such that $W(c)=0$.
5. No conclusion can be drawn without knowing $W$ is differentiable.

Explanation: The Intermediate Value Theorem (IVT) guarantees W continuous with W(-4) = -2 and W(-1) = 6 attains all values between -2 and 6 in (-4,-1). 0 is between them, making choice A correct. Choice B is wrong as 7 > 6, and choice C because -3 < -2. A common error is requiring differentiability for IVT, but continuity is enough, refuting choice E. Wrong intervals invalidate choice D. Transferable IVT checklist: 1. Confirm continuity on [a,b]. 2. Verify k is between f(a) and f(b). 3. Ensure c is in (a,b).

Question 15

1. A value $c\in(-1,2)$ such that $G(c)=5$.
2. A value $c\in(-1,2)$ such that $G(c)=1$. (correct answer)
3. A value $c\in(2,5)$ such that $G(c)=1$.
4. A value $c\in(-1,2)$ such that $G(c)=-3$.
5. A value $c$ is guaranteed only if $G$ is increasing.

Explanation: The Intermediate Value Theorem (IVT) applies to continuous G with G(-1) = -2 and G(2) = 4, guaranteeing values between -2 and 4. Since -2 < 1 < 4, there exists c in [-1, 2] with G(c) = 1, and as 1 ≠ -2, 1 ≠ 4, c is in (-1, 2). This works with mixed signs. A common error is choosing 5, outside the range. Requiring increasing behavior is unnecessary. Transferable IVT checklist: 1. Confirm the function is continuous on [a, b]. 2. Verify k is between f(a) and f(b). 3. If yes, there exists c in [a, b] with f(c) = k; if k is strictly between and ≠ f(a), ≠ f(b), then c is in (a, b).

Question 16

1. There exists $c\in(-3,1)$ such that $m(c)=8$.
2. There exists $c\in(-3,1)$ such that $m(c)=0$. (correct answer)
3. There exists $c\in(1,3)$ such that $m(c)=0$.
4. There exists $c\in(-3,1)$ such that $m(c)=-5$.
5. A value $c$ exists only if $m$ is linear.

Explanation: The Intermediate Value Theorem (IVT) applies to continuous m on [-3, 1] with m(-3) = -2 and m(1) = 6, guaranteeing values between -2 and 6. Since -2 < 0 < 6, there exists c in [-3, 1] with m(c) = 0, and as 0 ≠ -2, 0 ≠ 6, c is in (-3, 1). This is due to the sign change at endpoints. A common error is selecting values outside the range, like 8 or -5. Assuming linearity is needed is incorrect, as IVT works for any continuous function. Transferable IVT checklist: 1. Confirm the function is continuous on [a, b]. 2. Verify k is between f(a) and f(b). 3. If yes, there exists c in [a, b] with f(c) = k; if k is strictly between and ≠ f(a), ≠ f(b), then c is in (a, b).

Question 17

1. There exists $c\in(-1,2)$ such that $f(c)=3$. (correct answer)
2. There exists $c\in(-1,2)$ such that $f(c)=-1$.
3. There exists $c\in(-1,2)$ such that $f(c)=6$.
4. There exists $c\in(2,4)$ such that $f(c)=3$.
5. A root in $(-1,2)$ must exist.

Explanation: The Intermediate Value Theorem (IVT) for continuous f on [-1, 2] with f(-1) = 0 and f(2) = 5 guarantees values between 0 and 5. The value 3 is strictly between 0 and 5, so there exists c in [-1, 2] with f(c) = 3. Since 3 ≠ 0 and 3 ≠ 5, c is in (-1, 2). This highlights that IVT can guarantee non-zero values when appropriate. A common error is confusing this with guaranteeing a root, but 0 is already at an endpoint and not necessarily inside. Assuming the interval must be positive is irrelevant to IVT. Transferable IVT checklist: 1. Confirm the function is continuous on [a, b]. 2. Verify k is between f(a) and f(b). 3. If yes, there exists c in [a, b] with f(c) = k; if k is strictly between and ≠ f(a), ≠ f(b), then c is in (a, b).

Question 18

1. A value $c\in(-5,-1)$ such that $h(c)=10$.
2. A value $c\in(-5,-1)$ such that $h(c)=0$.
3. A value $c\in(-5,-1)$ such that $h(c)=5$. (correct answer)
4. A value $c\in(-1,0)$ such that $h(c)=5$.
5. No value is guaranteed because both endpoint values are positive.

Explanation: The Intermediate Value Theorem (IVT) applies to continuous h with h(-5) = 8 and h(-1) = 2, guaranteeing values between 2 and 8. Since 2 < 5 < 8, there exists c in [-5, -1] with h(c) = 5, and as 5 ≠ 8, 5 ≠ 2, c is in (-5, -1). This works despite same-sign endpoints because 5 is between them. A common error is assuming no guarantee without opposite signs, but IVT cares about the range. Thinking positive values prevent application is incorrect. Transferable IVT checklist: 1. Confirm the function is continuous on [a, b]. 2. Verify k is between f(a) and f(b). 3. If yes, there exists c in [a, b] with f(c) = k; if k is strictly between and ≠ f(a), ≠ f(b), then c is in (a, b).

Question 19

1. There exists $c\in(1,4)$ such that $N(c)=-9$. (correct answer)
2. There exists $c\in(1,4)$ such that $N(c)=0$.
3. There exists $c\in(1,4)$ such that $N(c)=-13$.
4. There exists $c\in(4,7)$ such that $N(c)=-9$.
5. No conclusion is possible unless $N$ is differentiable.

Explanation: The Intermediate Value Theorem (IVT) guarantees that continuous functions on [a,b] attain all values between f(a) and f(b) in (a,b). For N continuous with N(1) = -12 and N(4) = -6, values between -12 and -6 are assured in (1,4). -9 is between them, so choice A is correct. Choice C is wrong as -13 < -12, and choice B because 0 > -6. A common error is believing IVT needs differentiability, but continuity suffices, countering choice E. Applying to incorrect intervals misleads as in choice D. Transferable IVT checklist: 1. Confirm continuity on [a,b]. 2. Verify k is between f(a) and f(b). 3. Ensure c is in (a,b).

Question 20

1. A value $c\in(-2,3)$ such that $g(c)=7$.
2. A value $c\in(-2,3)$ such that $g(c)=0$. (correct answer)
3. A value $c\in(3,5)$ such that $g(c)=0$.
4. A value $c\in(-2,3)$ such that $g(c)=-3$.
5. No conclusion can be drawn without knowing $g'(x)$.

Explanation: The Intermediate Value Theorem (IVT) guarantees that a continuous function g on [-2, 3] takes every value between g(-2) = 6 and g(3) = -1. Since 6 > 0 > -1, 0 is between these values, so there exists c in [-2, 3] with g(c) = 0. As 0 ≠ g(-2) and 0 ≠ g(3), c is in (-2, 3). This demonstrates IVT's use in proving the existence of roots when endpoint signs differ. A common error is trying to apply IVT for values outside the range of g(-2) and g(3), like 7, which is not guaranteed. Another mistake is assuming differentiability is needed, but IVT requires only continuity. Transferable IVT checklist: 1. Confirm the function is continuous on [a, b]. 2. Verify k is between f(a) and f(b). 3. If yes, there exists c in [a, b] with f(c) = k; if k is strictly between and ≠ f(a), ≠ f(b), then c is in (a, b).