Answer: If $f$ is continuous on $[a, b]$ and $N$ is between $f(a)$ and $f(b)$, then there exists at least one $c \in (a, b)$ such that $f(c) = N$.

State the IVT precisely: The Intermediate Value Theorem says: if $f$ is continuous on $[a, b]$ and $N$ is any value strictly between $f(a)$ and $f(b)$, then there is at least one $c$ in $(a, b)$ with $f(c) = N$. Why the correct answer works: Option B is a direct statement of the IVT, capturing the hypothesis (continuity on a closed interval) and the conclusion (every intermediate value is attained). Why distractors fail: Option A describes the Extreme Value Theorem. Option C is the Mean Value Theorem. Option D is Rolle's Theorem.

Answer: The function $f(x) = 1/x$ is not continuous on $[-1, 1]$ because it is undefined at $x = 0$, so the IVT does not apply.

Verify the IVT hypotheses: The IVT requires $f$ to be continuous on the entire closed interval $[a, b]$. The function $f(x) = 1/x$ is not defined at $x = 0$, which lies in $[-1, 1]$, so it is not continuous on $[-1, 1]$. Identify the error: The student correctly evaluated the endpoints but failed to verify the hypothesis of continuity on the whole interval. Without this hypothesis, the IVT conclusion does not follow. Note the conceptual lesson: Indeed, $1/x$ never equals zero for any real $x$, which shows that skipping the continuity check can lead to false conclusions. Why distractors fail: Option A is wrong — the endpoint values are correctly computed. Option C is wrong — the IVT works regardless of the signs of $f(a)$ and $f(b)$. Option D is wrong — the IVT is commonly used to locate zeros.

Answer: Removable discontinuity

Check the continuity conditions: The limit $\lim_{x \to 3} f(x) = 5$ exists, and $f(3) = 7$ is defined, but $\lim_{x \to 3} f(x) \neq f(3)$. This means condition (3) for continuity fails. Classify the discontinuity: Because the two-sided limit exists but does not equal the function value, the discontinuity can be "removed" by redefining $f(3) = 5$. This is a removable discontinuity. Why distractors fail: A jump discontinuity requires different finite one-sided limits. An infinite discontinuity requires at least one side tending to $\pm\infty$. An oscillating discontinuity occurs when the function oscillates without settling to a limit.

Answer: $2a + b = 4$ only

Find the left-hand limit and function value: $\lim_{x \to 2^-} f(x) = 2^2 = 4$ and $f(2) = 4$. Find the right-hand limit: $\lim_{x \to 2^+} f(x) = a(2) + b = 2a + b$. Set up the continuity equation: For continuity at $x = 2$, we need $\lim_{x \to 2^+} f(x) = f(2)$, so $2a + b = 4$. This is the only condition — there is one equation in two unknowns, so infinitely many pairs $(a, b)$ work. Why distractors fail: Option B adds the unnecessary condition $a = 4$ (not required for continuity; differentiability would add a constraint, but continuity alone does not). Options C and D have incorrect equations.

Answer: $k = 1$

Find the limit at $x = 0$: The well-known limit $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ tells us the limiting behavior of $h$ near $x = 0$. Apply the continuity condition: For $h$ to be continuous at $x = 0$, we need $h(0) = \lim_{x \to 0} h(x)$, i.e., $k = 1$. Why distractors fail: Any value of $k$ other than $1$ would make $h(0) \neq \lim_{x \to 0} h(x)$, violating condition (3) for continuity.

Answer: Yes, by the Intermediate Value Theorem, since $22$ is between $18$ and $26$ and $T$ is continuous on $[0, 10]$.

Verify the IVT hypotheses: $T(t)$ is continuous on the closed interval $[0, 10]$, $T(0) = 18$, and $T(10) = 26$. The value $22$ lies between $18$ and $26$. Apply the IVT: By the IVT, there exists at least one $c \in (0, 10)$ such that $T(c) = 22$. No formula is needed — continuity alone suffices. Why distractors fail: Option A is wrong because the IVT does not require a formula, only continuity and endpoint values. Option B is wrong because differentiability is not a hypothesis of the IVT. Option C cites the wrong theorem; the EVT guarantees max/min, not intermediate values.

Answer: $g$ has a removable discontinuity at $x = 1$.

Compute the one-sided limits: $\lim_{x \to 1^-} g(x) = 2(1) + 1 = 3$ and $\lim_{x \to 1^+} g(x) = 1^2 + 2 = 3$. Since both one-sided limits equal $3$, $\lim_{x \to 1} g(x) = 3$. Compare with $g(1)$: $g(1) = 5 \neq 3 = \lim_{x \to 1} g(x)$. The limit exists, $g(1)$ is defined, but they are not equal. Classify: Because the two-sided limit exists but does not match the function value, the discontinuity is removable. Redefining $g(1) = 3$ would make $g$ continuous at $x = 1$. Why distractors fail: The function is not continuous (limit $\neq$ function value). It is not a jump (one-sided limits are equal). It is not infinite (all values are finite).

Answer: The IVT guarantees every value between $f(-2)$ and $f(5)$ is attained, but when $f(-2) = f(5)$ the interval of guaranteed values is empty, so the IVT says nothing about whether $f$ takes other values.

Understand the IVT conclusion: The IVT guarantees that every $N$ strictly between $f(a)$ and $f(b)$ is attained. When $f(a) = f(b) = 4$, there is no value strictly between them, so the IVT provides no information. Provide a counterexample to the student's claim: Consider $f(x) = (x - 1.5)^2 + 3$ adjusted so $f(-2) = f(5) = 4$. This function is not constant but satisfies the given conditions. Why distractors fail: Option A is wrong because the IVT still applies (the hypotheses are met); it just yields a trivial conclusion. Options C and D are factually incorrect — equal endpoint values do not force constancy.

Answer: $1$

Apply the IVT: Since $f$ is continuous on $[0, 4]$ with $f(0) = -2$ and $f(4) = 3$, the IVT guarantees that every value $N$ with $-2 < N < 3$ is attained by $f$ for some $c \in (0, 4)$. Check each option: $1$ lies in the interval $(-2, 3)$, so the IVT guarantees $f(c) = 1$ for some $c \in (0, 4)$. Why distractors fail: $-3 < -2$, so it is below $f(0)$ and not guaranteed. $4 > 3$ and $5 > 3$, so they exceed $f(4)$ and are not guaranteed. The function might attain these values, but the IVT does not guarantee it.

Answer: Being defined at every point ensures only condition (1) for continuity; the limit must also exist and equal the function value at each point.

Recall the three conditions: Continuity at a point requires (1) $f(a)$ is defined, (2) $\lim_{x \to a} f(x)$ exists, and (3) the limit equals $f(a)$. Being defined everywhere only guarantees condition (1). Construct a counterexample: Consider the piecewise function $f(x) = 1$ for $x \neq 0$ and $f(0) = 5$ on $[-1, 1]$. It is defined at every point, but at $x = 0$ the limit is $1 \neq 5 = f(0)$, so it is not continuous there. Why distractors fail: Option A is wrong because differentiability is not required for continuity. Option C is factually incorrect. Option D confuses continuity with the hypothesis of Rolle's Theorem.

Answer: $f(a)$ is defined, $\lim_{x \to a} f(x)$ exists, and $\lim_{x \to a} f(x) = f(a)$

Recall the three conditions for continuity: A function $f$ is continuous at $x = a$ if and only if: (1) $f(a)$ is defined, (2) $\lim_{x \to a} f(x)$ exists, and (3) $\lim_{x \to a} f(x) = f(a)$. Why the correct answer works: Option B states exactly these three conditions, with no additional or incorrect requirements. Why distractors fail: Options A, C, and D all include the requirement that $f'(a)$ exists. Differentiability is a stronger condition than continuity — a function can be continuous at a point without being differentiable there (e.g., $f(x) = |x|$ at $x = 0$).

Answer: $[0, 1]$

Evaluate $f$ at the endpoints of each interval: For $[0, 1]$: $f(0) = 1 > 0$ and $f(1) = 1 - 4 + 1 = -2 < 0$. Since $f$ changes sign, the IVT guarantees a root in $(0, 1)$. Check the other intervals: For $[2, 3]$: $f(2) = 8 - 8 + 1 = 1 > 0$ and $f(3) = 27 - 12 + 1 = 16 > 0$. No sign change. For $[-4, -3]$: $f(-4) = -64 + 16 + 1 = -47 < 0$ and $f(-3) = -27 + 12 + 1 = -14 < 0$. No sign change. For $[3, 4]$: $f(3) = 16 > 0$ and $f(4) = 64 - 16 + 1 = 49 > 0$. No sign change. Conclusion: Only on $[0, 1]$ do we have a sign change, which is required for the IVT to guarantee a root.

Answer: $f(x) = \frac{|x|}{x}$

Analyze $f(x) = |x|/x$: For $x > 0$, $|x|/x = 1$. For $x < 0$, $|x|/x = -1$. So $\lim_{x \to 0^+} f(x) = 1$ and $\lim_{x \to 0^-} f(x) = -1$. The one-sided limits are different finite values, so this is a jump discontinuity. Check the other options: $\sin(x)/x$ has limit $1$ as $x \to 0$ (removable discontinuity). $(x^2 - 1)/(x+1) = x - 1$ for $x \neq -1$, which is continuous at $x = 0$. $x\sin(1/x) \to 0$ as $x \to 0$ by the Squeeze Theorem (removable discontinuity). Why distractors fail: Options B and D have limits at $x = 0$ (removable, not jump). Option C has no discontinuity at $x = 0$ at all.

Answer: $f(x) = \begin{cases} x + 1 & \text{if } x \leq 1 \\ 2x - 2 & \text{if } x > 1 \end{cases}$

[{"title":"Check endpoint values for Option C","body":"$f(-1) = -1 + 1 = 0$ ✓ and $f(3) = 2(3) - 2 = 4$ ✓."},{"title":"Check for a jump discontinuity at $x = 1$","body":"$\\lim_{x \\to 1^-} f(x) = 1 + 1 = 2$ and $f(1) = 2$, but $\\lim_{x \\to 1^+} f(x) = 2(1) - 2 = 0$. Since $2 \\neq 0$, the one-sided limits differ, producing a jump discontinuity."},{"title":"Check that $f$ is defined on all of $[-1, 3]$","body":"Every $x \\in [-1, 3]$ falls into one of the two cases, so the function is defined everywhere on the interval."},{"title":"Why distractors fail","body":"Option A: $f(3) = 3 + 2 = 5 \\neq 4$. Option B: Both pieces are $x + 1$ — no discontinuity at all. "Option D: $f(-1) = 2(-1) + 3 = 1 \\neq 0$, so it fails the endpoint condition $f(-1) = 0$."}]

Answer: $f(x) = e^x$

Identify domains of each function: $\frac{1}{x-2}$ is undefined at $x=2$. $\ln(x)$ is defined only for $x > 0$. $\tan(x)$ is undefined at $x = \frac{\pi}{2} + n\pi$. The exponential function $e^x$ is defined and continuous on all of $(-\infty, \infty)$. Why the correct answer works: Exponential functions are continuous everywhere on their domain, which for $e^x$ is the entire real line. Why distractors fail: Each of the other three functions has points or intervals where it is either undefined or has discontinuities, so none is continuous for all real numbers.

Answer: $f$ has a removable discontinuity at $x = 3$, since the limit exists but $f(3)$ is undefined.

Simplify the expression: $f(x) = \frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x-3} = x + 3$ for $x \neq 3$. So $\lim_{x \to 3} f(x) = 6$. Check the continuity conditions: $f(3)$ is undefined because the original expression has a zero denominator at $x = 3$. The limit exists and equals $6$, but condition (1) fails. Classify the discontinuity: Since the limit exists but the function is not defined at the point, we can remove the discontinuity by defining $f(3) = 6$. This is a removable discontinuity. Why distractors fail: There is no jump (one-sided limits are equal) and no infinite behavior (the limit is finite). The function is not continuous because $f(3)$ is undefined.