AP Calculus BC Quiz: Working With The Intermediate Value Theorem
20 questions · exam conditions
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Working With The Intermediate Value TheoremQuestion 1 of 20

Let pp be continuous on [−1,2][-1,2] with p(−1)=−5p(-1)=-5 and p(2)=1p(2)=1. Does IVT guarantee some cc with p(c)=−3p(c)=-3?

No, because IVT applies only to polynomials.
Yes, because p(−1)p(-1) and p(2)p(2) have opposite signs.
Yes, because pp is continuous on [−1,2][-1,2] and −3-3 is between p(−1)p(-1) and p(2)p(2).
No, because −3-3 is not between −1-1 and 22.
Yes, because pp is differentiable on [−1,2][-1,2].
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AP Calculus BC Quiz

AP Calculus BC Quiz: Working With The Intermediate Value Theorem

Practice Working With The Intermediate Value Theorem in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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.

All questions

Question 1

Let pp be continuous on [−1,2][-1,2] with p(−1)=−5p(-1)=-5 and p(2)=1p(2)=1. Does IVT guarantee some cc with p(c)=−3p(c)=-3?

  1. No, because IVT applies only to polynomials.
  2. Yes, because p(−1)p(-1) and p(2)p(2) have opposite signs.
  3. Yes, because pp is continuous on [−1,2][-1,2] and −3-3 is between p(−1)p(-1) and p(2)p(2). (correct answer)
  4. No, because −3-3 is not between −1-1 and 22.
  5. Yes, because pp is differentiable on [−1,2][-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

Suppose hh is continuous on [0,6][0,6] with h(0)=10h(0)=10 and h(6)=10h(6)=10. Does IVT guarantee a solution to h(x)=0h(x)=0 on [0,6][0,6]?

  1. Yes, because hh is continuous on [0,6][0,6].
  2. No, because 00 is not between h(0)h(0) and h(6)h(6). (correct answer)
  3. Yes, because h(0)=h(6)h(0)=h(6).
  4. No, because hh may not be increasing on [0,6][0,6].
  5. Yes, because hh is continuous at x=0x=0 and x=6x=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

Let pp be continuous on [−1,5][-1,5] with p(−1)=−3p(-1)=-3 and p(5)=1p(5)=1; does IVT guarantee a solution to p(x)=2p(x)=2?

  1. Yes, because 22 is greater than the endpoint values.
  2. No, because 22 is not between the endpoint values −3-3 and 11. (correct answer)
  3. Yes, because 22 lies between −1-1 and 55.
  4. Yes, because p(−1)p(5)<0p(-1)p(5)<0 regardless of continuity.
  5. No, because IVT requires pp 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

A continuous function UU on [−2,6][-2,6] has U(−2)=9U(-2)=9 and U(6)=−3U(6)=-3; does IVT guarantee a solution to U(x)=7U(x)=7?

  1. No, because 77 is not between 99 and −3-3.
  2. Yes, because 77 lies between 99 and −3-3 and UU is continuous on [−2,6][-2,6]. (correct answer)
  3. Yes, because 77 lies between −2-2 and 66.
  4. No, because both endpoints are not integers.
  5. Yes, because U(−2)U(6)<0U(-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

If qq is continuous on [−1,1][-1,1] with q(−1)=0.2q(-1)=0.2 and q(1)=0.8q(1)=0.8, does IVT guarantee a solution to q(x)=0.5q(x)=0.5?

  1. No, because q(−1)q(-1) and q(1)q(1) are not integers.
  2. Yes, because 0.50.5 lies between q(−1)q(-1) and q(1)q(1) and qq is continuous on [−1,1][-1,1]. (correct answer)
  3. No, because IVT requires q(−1)eqq(1)q(-1) eq q(1) and here they are close.
  4. Yes, because qq is continuous at x=0x=0.
  5. No, because the interval must be [0,1][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

Function pp is continuous on [−1,4][-1,4] with p(−1)=3p(-1)=3 and p(4)=−6p(4)=-6; does IVT guarantee a solution to p(x)=−2p(x)=-2?

  1. No, because −2-2 is not between −1-1 and 44.
  2. Yes, because pp is continuous on [−1,4][-1,4] and −2-2 lies between 33 and −6-6. (correct answer)
  3. Yes, because p(−1)p(4)<0p(-1)p(4)<0 without needing continuity.
  4. No, because IVT applies only when endpoint values are integers.
  5. No, because pp might cross −2-2 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

A continuous function qq on [2,10][2,10] satisfies q(2)=5q(2)=5 and q(10)=−1q(10)=-1; does IVT guarantee a solution to q(x)=4q(x)=4?

  1. Yes, because 44 lies between 55 and −1-1 and qq is continuous on [2,10][2,10]. (correct answer)
  2. No, because 44 is not between 22 and 1010.
  3. No, because IVT requires opposite signs and 44 is positive.
  4. Yes, because qq is defined on (2,10)(2,10).
  5. Yes, because q(2)q(10)<0q(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

A continuous function uu on [4,5][4,5] has u(4)=−1u(4)=-1 and u(5)=−2u(5)=-2; does IVT guarantee a solution to u(x)=0u(x)=0?

  1. Yes, because 00 is greater than both endpoint values.
  2. No, because 00 is not between the endpoint values −1-1 and −2-2. (correct answer)
  3. Yes, because 00 lies between 44 and 55.
  4. Yes, because uu is continuous on [4,5][4,5].
  5. No, because IVT requires u(4)u(5)<0u(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

Let TT be continuous on [0,1][0,1] with T(0)=0T(0)=0 and T(1)=4T(1)=4; does IVT guarantee a solution to T(x)=3T(x)=3?

  1. Yes, because 33 lies between 00 and 44 and TT is continuous on [0,1][0,1]. (correct answer)
  2. No, because 33 is not between 00 and 11.
  3. Yes, because TT is defined on (0,1)(0,1).
  4. No, because IVT requires T(0)T(1)<0T(0)T(1)<0.
  5. Yes, because T(0)=0T(0)=0 forces T(x)=3T(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

A continuous function oo on [0,8][0,8] has o(0)=12o(0)=12 and o(8)=15o(8)=15; does IVT guarantee some cc with o(c)=14o(c)=14?

  1. Yes, because 1414 lies between 1212 and 1515 and oo is continuous on [0,8][0,8]. (correct answer)
  2. No, because 1414 is not between 00 and 88.
  3. No, because endpoint values must have opposite signs.
  4. Yes, because oo is defined at 00 and 88.
  5. Yes, because cc must equal 44.
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

A continuous function ss on [−7,−3][-7,-3] satisfies s(−7)=−2s(-7)=-2 and s(−3)=6s(-3)=6; does IVT guarantee a solution to s(x)=1s(x)=1?

  1. Yes, because 11 lies between −2-2 and 66 and ss is continuous on [−7,−3][-7,-3]. (correct answer)
  2. No, because 11 is not between −7-7 and −3-3.
  3. Yes, because s(−7)s(−3)<0s(-7)s(-3)<0 without continuity.
  4. No, because IVT applies only to s(x)=0s(x)=0.
  5. Yes, because 11 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

Suppose qq is continuous on [2,6][2,6], with q(2)=10q(2)=10 and q(6)=4q(6)=4; does IVT guarantee a solution to q(x)=7q(x)=7?​

  1. No, because 77 is not between the interval endpoints 2 and 6.
  2. Yes, because q(2)q(6)>0q(2)q(6)>0 ensures q(x)=7q(x)=7 for some xx.
  3. Yes, because qq is continuous on [2,6][2,6] and 77 is between q(2)q(2) and q(6)q(6). (correct answer)
  4. No, because IVT requires q(2)=−q(6)q(2)=-q(6) to guarantee q(x)=7q(x)=7.
  5. Yes, because qq is differentiable on (2,6)(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

If hh is continuous on [0,2][0,2], with h(0)=4h(0)=4 and h(2)=−1h(2)=-1, does IVT guarantee a solution to h(x)=2h(x)=2?​

  1. No, because h(0)h(2)<0h(0)h(2)<0 only guarantees a solution to h(x)=0h(x)=0.
  2. Yes, because hh is continuous on [0,2][0,2] and 22 is between h(0)h(0) and h(2)h(2). (correct answer)
  3. Yes, because 22 is between 0 and 2 and hh is defined at endpoints.
  4. No, because IVT requires h(0)=2h(0)=2 or h(2)=2h(2)=2.
  5. Yes, because hh is differentiable on (0,2)(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

Assume uu is continuous on [3,7][3,7] with u(3)=−1u(3)=-1 and u(7)=2u(7)=2; does IVT guarantee a solution to u(x)=5u(x)=5?​

  1. Yes, because uu is continuous on [3,7][3,7] and u(7)>0u(7)>0.
  2. No, because 55 is not between u(3)u(3) and u(7)u(7). (correct answer)
  3. Yes, because 55 is between 3 and 7 so u(x)=5u(x)=5 for some xx.
  4. No, because IVT requires u(3)u(7)<0u(3)u(7)<0 to guarantee u(x)=5u(x)=5.
  5. Yes, because uu is differentiable on (3,7)(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

Let rr be continuous on [−4,−1][-4,-1] with r(−4)=3r(-4)=3 and r(−1)=−6r(-1)=-6. Does IVT guarantee some cc with r(c)=−2r(c)=-2?

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

A continuous function gg on [−2,3][-2,3] satisfies g(−2)=4g(-2)=4 and g(3)=−1g(3)=-1. Does IVT guarantee some cc with g(c)=2g(c)=2?

  1. Yes, because gg is continuous on [−2,3][-2,3] and 22 is between g(−2)g(-2) and g(3)g(3). (correct answer)
  2. No, because 22 is not between −2-2 and 33.
  3. Yes, because g(−2)>g(3)g(-2)>g(3).
  4. No, because gg might not be differentiable on [−2,3][-2,3].
  5. Yes, because gg is defined at −2-2 and 33.
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

Let ff be continuous on [1,5][1,5] with f(1)=−2f(1)=-2 and f(5)=7f(5)=7. Does IVT guarantee a solution to f(x)=0f(x)=0?

  1. Yes, because ff is continuous on [1,5][1,5] and 00 is between f(1)f(1) and f(5)f(5). (correct answer)
  2. Yes, because f(1)f(1) and f(5)f(5) are defined on [1,5][1,5].
  3. No, because f(1)≠f(5)f(1)\ne f(5).
  4. Yes, because ff is continuous at x=1x=1 and x=5x=5.
  5. No, because IVT requires f(1)=0f(1)=0 or f(5)=0f(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

Let vv be continuous on [5,10][5,10] with v(5)=−9v(5)=-9 and v(10)=0v(10)=0. Does IVT guarantee some cc with v(c)=1v(c)=1?

  1. Yes, because vv is continuous on [5,10][5,10].
  2. Yes, because v(10)=0v(10)=0.
  3. No, because 11 is not between v(5)v(5) and v(10)v(10). (correct answer)
  4. Yes, because v(5)<v(10)v(5)<v(10).
  5. No, because vv might not be increasing on [5,10][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

Suppose ss is continuous on [0,2][0,2] with s(0)=−1s(0)=-1 and s(2)=5s(2)=5. Does IVT guarantee a solution to s(x)=7s(x)=7 on [0,2][0,2]?

  1. Yes, because ss is continuous on [0,2][0,2].
  2. No, because 77 is not between s(0)s(0) and s(2)s(2). (correct answer)
  3. Yes, because s(0)s(0) and s(2)s(2) have opposite signs.
  4. No, because ss might not be differentiable on [0,2][0,2].
  5. Yes, because ss is defined at 00 and 22.
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

Let tt be continuous on [3,8][3,8] with t(3)=12t(3)=12 and t(8)=−4t(8)=-4. Does IVT guarantee some cc with t(c)=6t(c)=6?

  1. No, because 66 is not between 33 and 88.
  2. Yes, because tt is continuous on [3,8][3,8] and 66 is between t(3)t(3) and t(8)t(8). (correct answer)
  3. Yes, because t(3)t(3) is positive.
  4. No, because t(8)t(8) is negative.
  5. Yes, because tt has values at x=3x=3 and x=8x=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.