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.

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.

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.

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 (✗).

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.

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).

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).

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.

This question tests whether IVT can guarantee a zero of q when both endpoint values are negative. The IVT applies when we have a continuous function on [a,b] and seek a value k that lies between f(a) and f(b). Here, q is continuous on [2,9] with q(2) = -8 and q(9) = -1, and we want q(x) = 0. To check if 0 is between the endpoint values: since both -8 and -1 are negative, we have -8 < -1 < 0, meaning 0 is not between q(2) and q(9). Choice C is incorrect because having both endpoint values negative actually prevents 0 from being between them—IVT cannot guarantee a root in this case. The IVT checklist: continuous function (✓), closed interval (✓), target between endpoints (✗ since 0 > both -8 and -1).

This problem evaluates understanding of the Intermediate Value Theorem (IVT), ensuring a continuous function hits every value between its endpoints on a closed interval. With h continuous on [0,3], h(0) = 5, and h(3) = 9, the values go from 5 to 9. Therefore, 7 is between 5 and 9, so IVT promises a c in [0,3] with h(c) = 7. This applies even if both endpoints are positive, as long as the target is in between. A distractor might wrongly insist on opposite signs, but IVT isn't limited to roots; it's for any intermediate value. Use this checklist for IVT: ensure closed-interval continuity and that k is between f(a) and f(b).