Working With the Intermediate Value Theorem
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AP Calculus BC › Working With the Intermediate Value Theorem
Let $p$ be continuous on $-1,2$ with $p(-1)=-5$ and $p(2)=1$. Does IVT guarantee some $c$ with $p(c)=-3$?
Yes, because $p$ is differentiable on $[-1,2]$.
No, because IVT applies only to polynomials.
No, because $-3$ is not between $-1$ and $2$.
Yes, because $p(-1)$ and $p(2)$ have opposite signs.
Yes, because $p$ is continuous on $[-1,2]$ and $-3$ is between $p(-1)$ and $p(2)$.
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.
Suppose $h$ is continuous on $0,6$ with $h(0)=10$ and $h(6)=10$. Does IVT guarantee a solution to $h(x)=0$ on $0,6$?
Yes, because $h(0)=h(6)$.
No, because $h$ may not be increasing on $[0,6]$.
No, because $0$ is not between $h(0)$ and $h(6)$.
Yes, because $h$ is continuous at $x=0$ and $x=6$.
Yes, because $h$ is continuous on $[0,6]$.
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 (✗).
Let $p$ be continuous on $-1,5$ with $p(-1)=-3$ and $p(5)=1$; does IVT guarantee a solution to $p(x)=2$?
No, because IVT requires $p$ to be differentiable.
No, because $2$ is not between the endpoint values $-3$ and $1$.
Yes, because $p(-1)p(5)<0$ regardless of continuity.
Yes, because $2$ lies between $-1$ and $5$.
Yes, because $2$ is greater than the endpoint values.
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.
A continuous function $U$ on $-2,6$ has $U(-2)=9$ and $U(6)=-3$; does IVT guarantee a solution to $U(x)=7$?
Yes, because $U(-2)U(6)<0$ without continuity.
Yes, because $7$ lies between $-2$ and $6$.
No, because $7$ is not between $9$ and $-3$.
No, because both endpoints are not integers.
Yes, because $7$ lies between $9$ and $-3$ and $U$ is continuous on $[-2,6]$.
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.
If $q$ is continuous on $-1,1$ with $q(-1)=0.2$ and $q(1)=0.8$, does IVT guarantee a solution to $q(x)=0.5$?
No, because $q(-1)$ and $q(1)$ are not integers.
Yes, because $q$ is continuous at $x=0$.
No, because IVT requires $q(-1) eq q(1)$ and here they are close.
No, because the interval must be $[0,1]$.
Yes, because $0.5$ lies between $q(-1)$ and $q(1)$ and $q$ is continuous on $[-1,1]$.
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).
Function $p$ is continuous on $-1,4$ with $p(-1)=3$ and $p(4)=-6$; does IVT guarantee a solution to $p(x)=-2$?
No, because $p$ might cross $-2$ more than once.
Yes, because $p(-1)p(4)<0$ without needing continuity.
No, because $-2$ is not between $-1$ and $4$.
Yes, because $p$ is continuous on $[-1,4]$ and $-2$ lies between $3$ and $-6$.
No, because IVT applies only when endpoint values are integers.
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.
A continuous function $q$ on $2,10$ satisfies $q(2)=5$ and $q(10)=-1$; does IVT guarantee a solution to $q(x)=4$?
Yes, because $4$ lies between $5$ and $-1$ and $q$ is continuous on $[2,10]$.
Yes, because $q(2)q(10)<0$ even if discontinuous.
Yes, because $q$ is defined on $(2,10)$.
No, because IVT requires opposite signs and $4$ is positive.
No, because $4$ is not between $2$ and $10$.
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.
A continuous function $u$ on $4,5$ has $u(4)=-1$ and $u(5)=-2$; does IVT guarantee a solution to $u(x)=0$?
No, because $0$ is not between the endpoint values $-1$ and $-2$.
No, because IVT requires $u(4)u(5)<0$ and differentiability.
Yes, because $u$ is continuous on $[4,5]$.
Yes, because $0$ lies between $4$ and $5$.
Yes, because $0$ is greater than both endpoint values.
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.
Let $T$ be continuous on $0,1$ with $T(0)=0$ and $T(1)=4$; does IVT guarantee a solution to $T(x)=3$?
No, because IVT requires $T(0)T(1)<0$.
Yes, because $3$ lies between $0$ and $4$ and $T$ is continuous on $[0,1]$.
Yes, because $T(0)=0$ forces $T(x)=3$ somewhere.
No, because $3$ is not between $0$ and $1$.
Yes, because $T$ is defined on $(0,1)$.
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.
A continuous function $o$ on $0,8$ has $o(0)=12$ and $o(8)=15$; does IVT guarantee some $c$ with $o(c)=14$?
No, because endpoint values must have opposite signs.
Yes, because $c$ must equal $4$.
No, because $14$ is not between $0$ and $8$.
Yes, because $o$ is defined at $0$ and $8$.
Yes, because $14$ lies between $12$ and $15$ and $o$ is continuous on $[0,8]$.
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.