The Intermediate Value Theorem (IVT) asserts that a continuous function on [a,b] takes every value between f(a) and f(b) inside (a,b). R is continuous on [-1,3] with R(-1) = 2 and R(3) = 10, guaranteeing values between 2 and 10. 6 is between them, making choice A correct. Choice B is incorrect as 12 > 10, and choice C because 0 < 2. A common error is thinking same-sign endpoints prevent IVT, but it applies to any value in the range, refuting 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) applies to the continuous function h on [0, 5], where h(0) = 1 and h(5) = 9, guaranteeing every value between 1 and 9 is achieved. The value 5 is strictly between 1 and 9, so there exists c in [0, 5] with h(c) = 5. Since 5 ≠ 1 and 5 ≠ 9, c must be in (0, 5). This shows how IVT ensures intermediate values are hit in the interior when strictly between endpoints. A common error is selecting a value outside the range, like 10 or -1, which IVT does not guarantee. Confusing IVT with the Mean Value Theorem by requiring differentiability is another mistake. 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) 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).

The Intermediate Value Theorem (IVT) ensures $A$ continuous on $[-1,1]$ with $A(-1) = 2$ and $A(1) = 6$ takes all values between 2 and 6 in $(-1,1)$. 4 is between them, so choice A is correct. Choice C fails as 7 > 6, and choice B because 0 < 2. A common error is assuming symmetry affects IVT, but it doesn't, 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 C continuous with C(-6) = -2 and C(-2) = -10 takes values between -10 and -2 in (-6,-2). -6 is between them, so choice A is correct. Choice C fails as -12 < -10, and choice B because 0 > -2. A common error is assuming same-sign endpoints prevent guarantees, but IVT applies to intermediates, refuting 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) 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) 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).

Given q continuous on [1,5] with q(1) = -3/2 and q(5) = 1/2, we need to find which value IVT guarantees. The function takes values from -3/2 to 1/2, and since these have opposite signs, 0 must be an intermediate value. This makes C the correct answer. Common errors include not recognizing that 0 always lies between a negative and positive value, or checking values like -2 or -3 which are outside the range [-3/2, 1/2]. The value 3/2 is also too large. IVT Checklist: ✓ Continuous function q, ✓ Endpoints have opposite signs, ✓ Zero is guaranteed between negative and positive values.

Given m continuous on [2,8] with m(2) = √2 ≈ 1.414 and m(8) = 5, IVT guarantees values between √2 and 5. Checking the options: 1/2 ≈ 0.5 is less than √2, while 6 exceeds 5, and √26 ≈ 5.099 is also greater than 5. Both √2 (option A) and 3 (option E) fall within the range, but option A is an endpoint value, not an intermediate value. Option E with value 3 is the only intermediate value in the range [√2, 5], making it correct. IVT Checklist: ✓ m is continuous, ✓ Range is [√2, 5] ≈ [1.414, 5], ✓ Value 3 is within this range.

Here p is continuous on [-1,2] with p(-1) = -4 and p(2) = -1. Even though both endpoint values are negative, IVT still applies to any value between -4 and -1. Since -2 is in the range [-4,-1], IVT guarantees there exists c in (-1,2) where p(c) = -2, making A correct. A common misconception is thinking IVT requires endpoint values to have opposite signs (for finding zeros) - but IVT works for any intermediate value regardless of signs. Option E incorrectly suggests no value is guaranteed, while -5 is outside the range. IVT Checklist: ✓ Continuous function, ✓ Both endpoints negative is fine, ✓ Target -2 is between -4 and -1.