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