AP Calculus BC 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 BC.

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. No, because IVT applies only to polynomials.
2. Yes, because $p(-1)$ and $p(2)$ have opposite signs.
3. Yes, because $p$ is continuous on $[-1,2]$ and $-3$ is between $p(-1)$ and $p(2)$. (correct answer)
4. No, because $-3$ is not between $-1$ and $2$.
5. Yes, because $p$ is differentiable on $[-1,2]$.

Explanation: This problem asks you to apply the Intermediate Value Theorem to determine if p(c) = -3 has a solution. The IVT guarantees that if a function is continuous on [a,b] and k is any value between f(a) and f(b), then f(c) = k for some c in [a,b]. Given that p is continuous on [-1,2] with p(-1) = -5 and p(2) = 1, we need to verify that -3 is between -5 and 1. Since -5 < -3 < 1, the value -3 is indeed between p(-1) and p(2), so IVT guarantees a solution exists. Choice D incorrectly thinks -3 needs to be between the x-values -1 and 2, but IVT requires the target value to be between the y-values at the endpoints. For IVT success: (1) verify continuity, (2) identify endpoint values, (3) check if target lies between them.

Question 2

1. Yes, because $h$ is continuous on $[0,6]$.
2. No, because $0$ is not between $h(0)$ and $h(6)$. (correct answer)
3. Yes, because $h(0)=h(6)$.
4. No, because $h$ may not be increasing on $[0,6]$.
5. Yes, because $h$ is continuous at $x=0$ and $x=6$.

Explanation: This question tests whether IVT can guarantee a root when the function has the same value at both endpoints. The IVT states that for a continuous function on [a,b], if k is between f(a) and f(b), then f(c) = k has a solution. Here, h is continuous on [0,6] with h(0) = 10 and h(6) = 10, and we want to know if h(x) = 0 has a solution. Since both endpoint values equal 10, the only value "between" h(0) and h(6) is 10 itself, so 0 is not between the endpoint values. Choice C is tempting because it correctly identifies that h(0) = h(6), but this equality doesn't help us apply IVT for finding where h(x) = 0. The IVT checklist requires: continuity (✓), closed interval (✓), but target value between endpoints (✗).

Question 3

1. Yes, because $2$ is greater than the endpoint values.
2. No, because $2$ is not between the endpoint values $-3$ and $1$. (correct answer)
3. Yes, because $2$ lies between $-1$ and $5$.
4. Yes, because $p(-1)p(5)<0$ regardless of continuity.
5. No, because IVT requires $p$ to be differentiable.

Explanation: This question demonstrates IVT limitations when the target value falls outside the range of endpoint values. The Intermediate Value Theorem states that if p is continuous on [a,b] and k lies between p(a) and p(b), then p(c) = k for some c in [a,b]. Here p is continuous on [-1,5], p(-1) = -3, and p(5) = 1. For p(c) = 2 to be guaranteed, 2 must lie between -3 and 1. Since 2 > 1, the value 2 does not lie between the endpoint values -3 and 1. Choice C incorrectly compares 2 with domain values. IVT checklist: (1) continuity on closed interval, (2) target value between endpoint function values.

Question 4

1. No, because $7$ is not between $9$ and $-3$.
2. Yes, because $7$ lies between $9$ and $-3$ and $U$ is continuous on $[-2,6]$. (correct answer)
3. Yes, because $7$ lies between $-2$ and $6$.
4. No, because both endpoints are not integers.
5. Yes, because $U(-2)U(6)<0$ without continuity.

Explanation: This problem applies the Intermediate Value Theorem when endpoint values have opposite signs. IVT guarantees that if U is continuous on [a,b] and k lies between U(a) and U(b), then U(c) = k for some c in [a,b]. Here U is continuous on [-2,6], U(-2) = 9, and U(6) = -3. Since 7 lies between -3 and 9, the IVT guarantees existence of c with U(c) = 7. Choice A incorrectly states that 7 is not between 9 and -3, but 7 is indeed between -3 and 9. For successful IVT application: (1) verify continuity on closed interval, (2) confirm target lies between endpoint function values.

Question 5

1. No, because $q(-1)$ and $q(1)$ are not integers.
2. Yes, because $0.5$ lies between $q(-1)$ and $q(1)$ and $q$ is continuous on $[-1,1]$. (correct answer)
3. No, because IVT requires $q(-1) eq q(1)$ and here they are close.
4. Yes, because $q$ is continuous at $x=0$.
5. No, because the interval must be $[0,1]$.

Explanation: This query tests Intermediate Value Theorem (IVT) application, stating that continuous functions on closed intervals achieve all intermediate values. Given q continuous on [-1,1] with q(-1) = 0.2 and q(1) = 0.8, 0.5 is between 0.2 and 0.8. So, IVT ensures a c in [-1,1] with q(c) = 0.5. The values don't need to be integers; IVT applies to all reals. One distractor might claim no because endpoints are close, but proximity doesn't affect the guarantee if the target is between. For IVT, verify: closed interval, continuity, and k between f(a) and f(b).

Question 6

1. No, because $-2$ is not between $-1$ and $4$.
2. Yes, because $p$ is continuous on $[-1,4]$ and $-2$ lies between $3$ and $-6$. (correct answer)
3. Yes, because $p(-1)p(4)<0$ without needing continuity.
4. No, because IVT applies only when endpoint values are integers.
5. No, because $p$ might cross $-2$ more than once.

Explanation: This question tests application of the Intermediate Value Theorem for finding intermediate values. IVT guarantees that if p is continuous on [a,b] and k lies between p(a) and p(b), then p(c) = k for some c in the interval. Given p is continuous on [-1,4], p(-1) = 3, and p(4) = -6, we check if -2 lies between these values. Since -2 is between -6 and 3, the IVT guarantees a solution to p(x) = -2. Choice A incorrectly states that -2 is not between -1 and 4, confusing input values with function output values. Apply IVT by confirming: (1) continuity on closed interval, (2) target value lies between endpoint function values.

Question 7

1. Yes, because $4$ lies between $5$ and $-1$ and $q$ is continuous on $[2,10]$. (correct answer)
2. No, because $4$ is not between $2$ and $10$.
3. No, because IVT requires opposite signs and $4$ is positive.
4. Yes, because $q$ is defined on $(2,10)$.
5. Yes, because $q(2)q(10)<0$ even if discontinuous.

Explanation: This problem tests IVT application when endpoint values have opposite signs and the target lies between them. The Intermediate Value Theorem guarantees that if q is continuous on [a,b] and k lies between q(a) and q(b), then q(c) = k for some c in [a,b]. Given q is continuous on [2,10], q(2) = 5, and q(10) = -1, we check if 4 lies between these endpoint values. Since -1 < 4 < 5, the value 4 lies between the endpoint values, so IVT guarantees a solution to q(x) = 4. Choice B incorrectly compares the target with domain endpoints rather than function values. Apply IVT by confirming: (1) continuity on closed interval, (2) target between endpoint function values.

Question 8

1. Yes, because $0$ is greater than both endpoint values.
2. No, because $0$ is not between the endpoint values $-1$ and $-2$. (correct answer)
3. Yes, because $0$ lies between $4$ and $5$.
4. Yes, because $u$ is continuous on $[4,5]$.
5. No, because IVT requires $u(4)u(5)<0$ and differentiability.

Explanation: This problem demonstrates IVT limitations when the target value falls outside the range of endpoint values. The Intermediate Value Theorem guarantees that if u is continuous on [a,b] and k lies between u(a) and u(b), then u(c) = k for some c in [a,b]. Here u is continuous on [4,5], u(4) = -1, and u(5) = -2. For u(c) = 0 to be guaranteed, 0 must lie between -1 and -2. Since 0 > -1 > -2, the value 0 does not lie between the endpoint values -2 and -1. Choice D incorrectly suggests continuity alone is sufficient. IVT checklist: (1) continuity on closed interval, (2) target value between endpoint function values.

Question 9

1. Yes, because $3$ lies between $0$ and $4$ and $T$ is continuous on $[0,1]$. (correct answer)
2. No, because $3$ is not between $0$ and $1$.
3. Yes, because $T$ is defined on $(0,1)$.
4. No, because IVT requires $T(0)T(1)<0$.
5. Yes, because $T(0)=0$ forces $T(x)=3$ somewhere.

Explanation: This question tests IVT application when both endpoint values are positive and the target lies between them. The Intermediate Value Theorem states that if T is continuous on [a,b] and k lies between T(a) and T(b), then T(c) = k for some c in [a,b]. Given T is continuous on [0,1], T(0) = 0, and T(1) = 4, we need 3 to lie between these endpoint values. Since 0 < 3 < 4, the value 3 lies between the endpoint values, so IVT guarantees a solution to T(x) = 3. Choice B incorrectly compares the target with domain values rather than function values. IVT checklist: (1) continuity on closed interval, (2) target value between endpoint function values.

Question 10

1. Yes, because $14$ lies between $12$ and $15$ and $o$ is continuous on $[0,8]$. (correct answer)
2. No, because $14$ is not between $0$ and $8$.
3. No, because endpoint values must have opposite signs.
4. Yes, because $o$ is defined at $0$ and $8$.
5. Yes, because $c$ must equal $4$.

Explanation: This problem applies the Intermediate Value Theorem when both endpoint values are positive and the target lies between them. IVT guarantees that if o is continuous on [a,b] and k lies between o(a) and o(b), then o(c) = k for some c in [a,b]. Here o is continuous on [0,8], o(0) = 12, and o(8) = 15. Since 14 lies between 12 and 15, the IVT guarantees existence of c with o(c) = 14. Choice B incorrectly compares the target value 14 with the domain endpoints 0 and 8 rather than the function values. For IVT application: (1) verify continuity on closed interval, (2) confirm target lies between endpoint function values.

Question 11

1. Yes, because $1$ lies between $-2$ and $6$ and $s$ is continuous on $[-7,-3]$. (correct answer)
2. No, because $1$ is not between $-7$ and $-3$.
3. Yes, because $s(-7)s(-3)<0$ without continuity.
4. No, because IVT applies only to $s(x)=0$.
5. Yes, because $1$ is positive.

Explanation: This problem applies the Intermediate Value Theorem when endpoint values have opposite signs and the target lies between them. IVT guarantees that if s is continuous on [a,b] and k lies between s(a) and s(b), then s(c) = k for some c in [a,b]. Given s is continuous on [-7,-3], s(-7) = -2, and s(-3) = 6, we need 1 to lie between these endpoint values. Since -2 < 1 < 6, the value 1 lies between the endpoint values, so IVT guarantees a solution to s(x) = 1. Choice B incorrectly compares the target with domain endpoints rather than function values. IVT requires: (1) continuity on closed interval, (2) target value between endpoint function values.

Question 12

1. No, because $7$ is not between the interval endpoints 2 and 6.
2. Yes, because $q(2)q(6)>0$ ensures $q(x)=7$ for some $x$.
3. Yes, because $q$ is continuous on $[2,6]$ and $7$ is between $q(2)$ and $q(6)$. (correct answer)
4. No, because IVT requires $q(2)=-q(6)$ to guarantee $q(x)=7$.
5. Yes, because $q$ is differentiable on $(2,6)$, so it must hit 7.

Explanation: This question tests whether you can correctly apply the Intermediate Value Theorem to guarantee a solution exists. Since q is continuous on [2,6] with q(2) = 10 and q(6) = 4, the IVT ensures q attains every value between 4 and 10. The target value 7 lies within this range, so there must be at least one x in [2,6] where q(x) = 7. Choice A incorrectly focuses on whether 7 is between the x-values 2 and 6, but IVT concerns whether 7 is between the y-values q(2) and q(6). To apply IVT: check continuity on the interval, verify the target lies between endpoint function values, then conclude a solution exists.

Question 13

1. No, because $h(0)h(2)<0$ only guarantees a solution to $h(x)=0$.
2. Yes, because $h$ is continuous on $[0,2]$ and $2$ is between $h(0)$ and $h(2)$. (correct answer)
3. Yes, because $2$ is between 0 and 2 and $h$ is defined at endpoints.
4. No, because IVT requires $h(0)=2$ or $h(2)=2$.
5. Yes, because $h$ is differentiable on $(0,2)$ and crosses every value.

Explanation: This question tests your ability to apply the Intermediate Value Theorem when the target value lies between the endpoint function values. Since h is continuous on [0,2] with h(0) = 4 and h(2) = -1, the IVT guarantees that h takes on every value between -1 and 4. The target value 2 lies in this range, so there must be at least one c in [0,2] where h(c) = 2. Choice C incorrectly suggests that having 2 between the x-coordinates 0 and 2 matters, but IVT is about y-values, not x-values. For IVT success: verify continuity, confirm the target lies between endpoint function values, and conclude existence.

Question 14

1. Yes, because $u$ is continuous on $[3,7]$ and $u(7)>0$.
2. No, because $5$ is not between $u(3)$ and $u(7)$. (correct answer)
3. Yes, because $5$ is between 3 and 7 so $u(x)=5$ for some $x$.
4. No, because IVT requires $u(3)u(7)<0$ to guarantee $u(x)=5$.
5. Yes, because $u$ is differentiable on $(3,7)$ so it reaches 5.

Explanation: This question tests recognition of when the Intermediate Value Theorem does NOT guarantee a solution. Although u is continuous on [3,7] with u(3) = -1 and u(7) = 2, the IVT only guarantees u takes on values between -1 and 2. The target value 5 exceeds this range (5 > 2), so IVT cannot guarantee a solution to u(x) = 5. Choice C incorrectly focuses on 5 being between the x-values 3 and 7, but IVT requires the target to be between the y-values u(3) and u(7). Remember: IVT guarantees existence only when the target lies between (or equals) the endpoint function values.

Question 15

1. No, because $-2$ is not between $-4$ and $-1$.
2. Yes, because $r$ is continuous on $[-4,-1]$ and $-2$ is between $r(-4)$ and $r(-1)$. (correct answer)
3. Yes, because $r(-4)>0$.
4. No, because IVT requires $r(-4)=0$.
5. Yes, because $r$ is continuous at $x=-4$ and $x=-1$.

Explanation: This problem requires applying the Intermediate Value Theorem to find if r(c) = -2 has a solution. The IVT states that for a continuous function on [a,b], any value k between f(a) and f(b) is attained by the function. Given r is continuous on [-4,-1] with r(-4) = 3 and r(-1) = -6, we check if -2 is between 3 and -6. Since -6 < -2 < 3, the value -2 lies between r(-4) and r(-1), so IVT guarantees there exists c in [-4,-1] where r(c) = -2. Choice A incorrectly compares -2 with the domain values -4 and -1, but IVT requires comparing with the range values at the endpoints. Remember: IVT needs (1) continuity on [a,b], (2) target value between f(a) and f(b), not between a and b.

Question 16

1. Yes, because $g$ is continuous on $[-2,3]$ and $2$ is between $g(-2)$ and $g(3)$. (correct answer)
2. No, because $2$ is not between $-2$ and $3$.
3. Yes, because $g(-2)>g(3)$.
4. No, because $g$ might not be differentiable on $[-2,3]$.
5. Yes, because $g$ is defined at $-2$ and $3$.

Explanation: This problem requires applying the Intermediate Value Theorem to find if g(c) = 2 has a solution. The IVT applies when a function is continuous on a closed interval [a,b] and we're looking for a value k that lies between f(a) and f(b). Given that g is continuous on [-2,3] with g(-2) = 4 and g(3) = -1, we need to check if 2 is between 4 and -1. Since -1 < 2 < 4, the value 2 is indeed between g(-2) and g(3), so IVT guarantees there exists at least one c in [-2,3] where g(c) = 2. Choice B incorrectly confuses the domain interval [-2,3] with the range values—the target value 2 needs to be between the function values, not the x-values. Remember the IVT checklist: continuous function, closed interval, and target value between endpoint function values.

Question 17

1. Yes, because $f$ is continuous on $[1,5]$ and $0$ is between $f(1)$ and $f(5)$. (correct answer)
2. Yes, because $f(1)$ and $f(5)$ are defined on $[1,5]$.
3. No, because $f(1)\ne f(5)$.
4. Yes, because $f$ is continuous at $x=1$ and $x=5$.
5. No, because IVT requires $f(1)=0$ or $f(5)=0$.

Explanation: This question tests your ability to apply the Intermediate Value Theorem (IVT) to determine if a function has a root. The IVT states that if f is continuous on [a,b] and k is any value between f(a) and f(b), then there exists at least one c in [a,b] where f(c) = k. Here, f is continuous on [1,5], f(1) = -2, and f(5) = 7, so we need to check if 0 is between -2 and 7. Since -2 < 0 < 7, the value 0 is indeed between f(1) and f(5), so IVT guarantees there exists some c in [1,5] where f(c) = 0. Choice B is incorrect because merely having defined values at the endpoints isn't sufficient—we need continuity and the target value must be between the endpoint values. When applying IVT, always verify: (1) continuity on the closed interval, (2) the target value lies between the function values at the endpoints.

Question 18

1. Yes, because $v$ is continuous on $[5,10]$.
2. Yes, because $v(10)=0$.
3. No, because $1$ is not between $v(5)$ and $v(10)$. (correct answer)
4. Yes, because $v(5)<v(10)$.
5. No, because $v$ might not be increasing on $[5,10]$.

Explanation: This problem requires determining whether IVT guarantees v(c) = 1 when one endpoint is zero. The IVT applies when a continuous function on [a,b] takes a value k that lies between f(a) and f(b). Given v is continuous on [5,10] with v(5) = -9 and v(10) = 0, we check if 1 is between these values. Since -9 < 0 < 1, we see that 1 is not between v(5) and v(10)—it exceeds both endpoint values. Choice B incorrectly thinks having v(10) = 0 helps guarantee v(c) = 1, but the zero at an endpoint doesn't extend the range of guaranteed values beyond [v(5), v(10)]. Remember the IVT checklist: continuous function (✓), closed interval (✓), target between endpoints (✗ since 1 > 0 > -9).

Question 19

1. Yes, because $s$ is continuous on $[0,2]$.
2. No, because $7$ is not between $s(0)$ and $s(2)$. (correct answer)
3. Yes, because $s(0)$ and $s(2)$ have opposite signs.
4. No, because $s$ might not be differentiable on $[0,2]$.
5. Yes, because $s$ is defined at $0$ and $2$.

Explanation: This question tests whether IVT guarantees a solution to s(x) = 7 when the target exceeds both endpoint values. The IVT applies when a continuous function on [a,b] takes on any value k between f(a) and f(b). Here, s is continuous on [0,2] with s(0) = -1 and s(2) = 5, and we seek s(x) = 7. To check if 7 is between the endpoint values: we have -1 < 5 < 7, so 7 is not between s(0) and s(2)—it exceeds both values. Choice C incorrectly focuses on the signs of the endpoint values, but having opposite signs only helps when seeking a zero, not when seeking 7. The IVT checklist requires: continuous function (✓), closed interval (✓), but target between endpoints (✗ since 7 > 5 > -1).

Question 20

1. No, because $6$ is not between $3$ and $8$.
2. Yes, because $t$ is continuous on $[3,8]$ and $6$ is between $t(3)$ and $t(8)$. (correct answer)
3. Yes, because $t(3)$ is positive.
4. No, because $t(8)$ is negative.
5. Yes, because $t$ has values at $x=3$ and $x=8$.

Explanation: This problem asks you to apply the Intermediate Value Theorem to determine if t(c) = 6 has a solution. The IVT guarantees that a continuous function on [a,b] attains every value between f(a) and f(b). Given t is continuous on [3,8] with t(3) = 12 and t(8) = -4, we need to verify that 6 is between these endpoint values. Since -4 < 6 < 12, the value 6 is indeed between t(3) and t(8), so IVT guarantees there exists at least one c in [3,8] where t(c) = 6. Choice A incorrectly thinks 6 needs to be between the x-values 3 and 8, but IVT compares the target with the y-values at endpoints. For successful IVT application: (1) confirm continuity, (2) find endpoint function values, (3) verify target lies between them.