The Intermediate Value Theorem (IVT) ensures J continuous on [-1,5] with J(-1) = 3 and J(5) = -9 attains all values between -9 and 3 in (-1,5). 0 is between them, so choice B is correct. Choice C fails as 6 > 3, and choice A because -12 < -9. A common error is requiring a polynomial for IVT, but continuity suffices, refuting choice E. Wrong intervals mislead in choice D. Transferable IVT checklist: 1. Confirm continuity on [a,b]. 2. Verify k is between f(a) and f(b). 3. Ensure c is in (a,b).

The Intermediate Value Theorem (IVT) guarantees that a continuous function g on [-2, 3] takes every value between g(-2) = 6 and g(3) = -1. Since 6 > 0 > -1, 0 is between these values, so there exists c in [-2, 3] with g(c) = 0. As 0 ≠ g(-2) and 0 ≠ g(3), c is in (-2, 3). This demonstrates IVT's use in proving the existence of roots when endpoint signs differ. A common error is trying to apply IVT for values outside the range of g(-2) and g(3), like 7, which is not guaranteed. Another mistake is assuming differentiability is needed, but IVT requires only continuity. Transferable IVT checklist: 1. Confirm the function is continuous on [a, b]. 2. Verify k is between f(a) and f(b). 3. If yes, there exists c in [a, b] with f(c) = k; if k is strictly between and ≠ f(a), ≠ f(b), then c is in (a, b).

The Intermediate Value Theorem (IVT) ensures continuous f with f(0) = -1 and f(1) = 2 takes values between -1 and 2. Since -1 < 0 < 2, there exists c in [0, 1] with f(c) = 0, and as 0 ≠ -1, 0 ≠ 2, c is in (0, 1). This is a root guarantee from opposite signs. A common error is choosing 3, outside the range. Unequal endpoints do not prevent application; they enable it here. Transferable IVT checklist: 1. Confirm the function is continuous on [a, b]. 2. Verify k is between f(a) and f(b). 3. If yes, there exists c in [a, b] with f(c) = k; if k is strictly between and ≠ f(a), ≠ f(b), then c is in (a, b).

With s continuous on [0,2] where s(0) = -1 and s(2) = 4, IVT guarantees any value between -1 and 4. Examining the options: 5 and 6 exceed 4, while -2 and -3 are less than -1. Only option B with value 3 falls within the range [-1,4], making it the guaranteed value. Students often make errors by not carefully comparing each option to the endpoint values, or by assuming IVT might work for values just outside the range. IVT Checklist: ✓ s continuous on closed interval, ✓ Range is [-1,4], ✓ Check each option against this range.

Given that h is continuous on [0,3] with h(0) = 1/2 and h(3) = 7/2, we need to identify which value is guaranteed by IVT. The function values at the endpoints create the range [1/2, 7/2]. Any value k in this range must have at least one c in (0,3) where h(c) = k. Among the options, only 2 falls within [1/2, 7/2], making C the correct answer. Students often err by not carefully checking whether values fall within the endpoint range - for instance, 0 and 9/2 are outside this range. IVT Checklist: ✓ Continuous on closed interval [0,3], ✓ Identify range [1/2, 7/2], ✓ Check which option falls in this range.

The function q is continuous on [3,7] with q(3) = -1 and q(7) = -9. Both endpoint values are negative, with -9 < -1. By IVT, q must attain every value between -9 and -1 on the interval (3,7). The value -5 is between -9 and -1, so there exists c in (3,7) such that q(c) = -5. Values like 0 and 5 are not between the endpoint values (both are greater than -1), so IVT doesn't guarantee them. A common error is assuming IVT applies when both endpoints have the same sign but forgetting to check if the target value is actually between them. IVT Checklist: (1) Continuous on closed interval, (2) Target between min and max of endpoint values, (3) Correct interval in conclusion.

For function F continuous on [2,6] with F(2) = 10 and F(6) = 3, IVT guarantees the existence of any value between 3 and 10. Looking at the options: 2 and -1 are too small (less than 3), 11 is too large (greater than 10), and 0 is also outside the range [3,10]. Only option B with value 7 falls within [3,10], so IVT guarantees there exists c in (2,6) where F(c) = 7. Students sometimes forget that IVT only guarantees values strictly between the endpoint function values. IVT Checklist: ✓ F continuous on [2,6], ✓ Range is [3,10], ✓ Only 7 falls in this range.

Given that g is continuous on an interval containing [1,6] with g(1) = -4 and g(6) = 2, we can apply IVT. Since g changes from negative to positive, and 0 is between -4 and 2, IVT guarantees there exists c in (1,6) such that g(c) = 0. The value -5 in option B is not between the endpoint values, so IVT doesn't guarantee it exists. Similarly, we cannot conclude anything about intervals like (6,10) or (1,2) without knowing the function values at those specific endpoints. A common mistake is assuming IVT works for values outside the range of the given endpoints. IVT Checklist: (1) Verify continuity, (2) Check if target value is between endpoint values, (3) Use the correct interval in the conclusion.

The Intermediate Value Theorem (IVT) guarantees K continuous with K(3) = 2 and K(11) = -6 takes values between -6 and 2 in (3,11). 0 is between them, making choice A correct. Choice B is wrong as 3 > 2, and choice C because -8 < -6. A common error is thinking a positive start prevents roots, but sign change ensures it, countering choice E. Wrong intervals invalidate choice D. Transferable IVT checklist: 1. Confirm continuity on [a,b]. 2. Verify k is between f(a) and f(b). 3. Ensure c is in (a,b).

The Intermediate Value Theorem (IVT) states that for a continuous function f on a closed interval [a, b], any value k between f(a) and f(b) is achieved at some c in [a, b]. Here, f(1) = -3 and f(4) = 2, with -3 < 0 < 2, so IVT guarantees a c in [1, 4] where f(c) = 0. Since 0 ≠ f(1) and 0 ≠ f(4), this c must be in the open interval (1, 4). This application shows that the function must cross the x-axis between 1 and 4 due to the sign change. A common error is assuming IVT applies without a sign change or without continuity, but here both conditions are met. Another mistake is thinking IVT guarantees exactly one root, whereas it only ensures at least one. Transferable IVT checklist: 1. Confirm the function is continuous on [a, b]. 2. Verify k is between f(a) and f(b). 3. If yes, there exists c in [a, b] with f(c) = k; if k is strictly between and ≠ f(a), ≠ f(b), then c is in (a, b).