Answer: $\dfrac{f(b) - f(a)}{b - a}$

State the MVT conclusion: The Mean Value Theorem states: if $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists $c \in (a, b)$ such that $f'(c) = \dfrac{f(b) - f(a)}{b - a}$. Why the correct answer works: The expression $\dfrac{f(b) - f(a)}{b - a}$ is the average rate of change (slope of the secant line), which the MVT guarantees equals the instantaneous rate of change at some interior point. Why distractors fail: Option A gives the average of the function values, not the average rate of change. Option C uses a product of function values, which has no role in the MVT. Option D is the total change in $f$, missing the division by the interval length.

Answer: Yes — $\cos(c) = 0$ has only one solution in $(0, \pi)$, which is $c = \pi/2$.

Compute the average rate of change: $f(0) = 0$ and $f(\pi) = 0$, so $\dfrac{f(\pi)-f(0)}{\pi - 0} = 0$. Solve $f'(c) = 0$: $f'(x) = \cos(x)$. Setting $\cos(c) = 0$ on $(0, \pi)$ gives $c = \pi/2$ as the only solution. (The next zero of cosine is at $3\pi/2$, outside the interval.) Evaluate the claim: The student's conclusion is correct: $c = \pi/2$ is indeed the unique value satisfying the MVT conclusion on this interval. Why distractors fail: Option B is wrong because $\cos(c) = 0$ on $(0, \pi)$ yields only $c = \pi/2$. Option C is wrong because the MVT applies to any function meeting continuity and differentiability conditions, not only polynomials. Option D is wrong because while Rolle's Theorem also applies (since $f(0)=f(\pi)$), using the MVT is perfectly valid and gives the same result.

Answer: $c = 2$

Verify the hypotheses: $f(x) = x^2 - 4x + 3$ is a polynomial, so it is continuous on $[1,3]$ and differentiable on $(1,3)$. The MVT applies. Compute the average rate of change: $f(1) = 1 - 4 + 3 = 0$ and $f(3) = 9 - 12 + 3 = 0$. So $\dfrac{f(3)-f(1)}{3-1} = \dfrac{0-0}{2} = 0$. Solve for c: $f'(x) = 2x - 4$. Setting $f'(c) = 0$ gives $2c - 4 = 0$, so $c = 2$. Since $2 \in (1,3)$, this is valid. Why distractors fail: Option A ($c=1$) and Option C ($c=3$) are endpoints, and the MVT guarantees $c$ in the open interval. Option D ($c = 5/2$) does not satisfy $f'(c) = 0$.

Answer: $f$ is not differentiable at $x = 0$, which is in the open interval $(-1, 1)$.

Check the MVT hypotheses: $f(x) = |x|$ is continuous on $[-1, 1]$, so continuity holds. However, $f$ has a corner at $x = 0$, so $f'(0)$ does not exist. Identify the failing hypothesis: The MVT requires differentiability on the open interval $(-1, 1)$. Since $f$ is not differentiable at $x = 0 \in (-1, 1)$, this hypothesis fails. Why distractors fail: Option A is wrong because $|x|$ is continuous everywhere. Option C is wrong because $f(-1) = 1 = f(1)$, so the values are equal (and this is not even a hypothesis of the MVT, only of Rolle's). Option D is wrong because $|x|$ is bounded on any closed interval.

Answer: The function must satisfy $f(a) = f(b)$.

Recall the hypotheses of Rolle's Theorem: Rolle's Theorem requires three conditions: (1) $f$ is continuous on $[a, b]$, (2) $f$ is differentiable on $(a, b)$, and (3) $f(a) = f(b)$. Why the correct answer works: The condition $f(a) = f(b)$ is one of the three required hypotheses. Without it, Rolle's Theorem does not apply. Why distractors fail: Option A is incorrect because differentiability is required on the open interval $(a, b)$, not the closed interval. Option C is wrong because Rolle's Theorem does not require the function to be increasing. Option D is wrong because there is no requirement about critical points at the endpoints.

Answer: The tangent line at some interior point is parallel to the secant line through the endpoints.

Interpret MVT geometrically: The secant line through $(a, f(a))$ and $(b, f(b))$ has slope $\dfrac{f(b)-f(a)}{b-a}$. The MVT says the tangent line at $x = c$ has the same slope, meaning they are parallel. Why the correct answer works: A tangent line parallel to the secant line is the standard geometric statement of the MVT. Why distractors fail: Option A describes a concept closer to the Mean Value Theorem for Integrals, not the derivative form. Option C describes something related to the Intermediate Value Theorem. Option D describes a property that is not generally true and is unrelated to the MVT.

Answer: $f$ is a constant function on $[0, 5]$.

Apply the MVT consequence for zero derivatives: If $f'(x) = 0$ for all $x$ in an interval, then for any two points $x_1, x_2$ in that interval, the MVT gives $f(x_2) - f(x_1) = f'(c)(x_2 - x_1) = 0$, so $f(x_1) = f(x_2)$. Why the correct answer works: Since any two outputs are equal, $f$ is constant on $[0, 5]$. Why distractors fail: Option A is wrong because a zero derivative means the function is flat, not increasing. Option C is wrong because every point is a critical point when $f'=0$ everywhere. Option D is wrong because the function is constant but not necessarily equal to zero — it could be any constant.

Answer: $13$

Apply the MVT to bound the function: By the MVT, $f(3) - f(0) = f'(c)(3 - 0) = 3f'(c)$ for some $c \in (0, 3)$. Since $f'(c) \leq 4$, we get $f(3) - 1 \leq 3 \cdot 4 = 12$. Compute the bound: $f(3) \leq 1 + 12 = 13$. This maximum is achieved by $f(x) = 4x + 1$, which has $f'(x) = 4$ everywhere. Why distractors fail: Option A ($12$) neglects adding the initial value $f(0) = 1$. Option C ($4$) uses only the bound on $f'$ without multiplying by the interval length. Option D ($7$) may result from using the interval length incorrectly (e.g., multiplying $4 \times 1.5$).

Answer: Rolle's Theorem is the MVT with the additional condition $f(a) = f(b)$, so the guaranteed slope is zero.

Compare MVT and Rolle's Theorem: The MVT guarantees $f'(c) = \dfrac{f(b)-f(a)}{b-a}$. When $f(a) = f(b)$, this fraction equals zero, so $f'(c) = 0$. That is exactly Rolle's Theorem. Why the correct answer works: Option B correctly identifies the extra condition ($f(a) = f(b)$) and the resulting conclusion ($f'(c) = 0$). Why distractors fail: Option A is wrong because Rolle's Theorem still requires differentiability on $(a, b)$. Option C is wrong because Rolle's Theorem requires both continuity and differentiability. Option D states an incorrect conclusion — the guaranteed value is $f'(c) = 0$, not $f'(c) = f(a)$.

Answer: $f(a) < f(b)$

Apply the MVT consequence for positive derivatives: By the MVT, $f(b) - f(a) = f'(c)(b - a)$ for some $c \in (a, b)$. Since $f'(c) > 0$ and $b - a > 0$, we get $f(b) - f(a) > 0$. Why the correct answer works: $f(b) - f(a) > 0$ means $f(a) < f(b)$, confirming $f$ is strictly increasing from $a$ to $b$. Why distractors fail: Option A reverses the inequality. Option B claims equality, contradicting the positive derivative. Option D is wrong because a strictly increasing function on an interval has no local maximum in the interior.

Answer: $c = 1$

Compute the average rate of change: $f(0) = 0$ and $f(4) = 2$. The average rate of change is $\dfrac{2 - 0}{4 - 0} = \dfrac{1}{2}$. Find the derivative and solve: $f'(x) = \dfrac{1}{2\sqrt{x}}$. Setting $\dfrac{1}{2\sqrt{c}} = \dfrac{1}{2}$ gives $\sqrt{c} = 1$, so $c = 1$. Verify c is in the interval: $c = 1 \in (0, 4)$, so this value is valid. Note that $f$ is continuous on $[0,4]$ and differentiable on $(0,4)$. Why distractors fail: Option A ($c=2$): $f'(2) = \frac{1}{2\sqrt{2}} \neq \frac{1}{2}$. Option C ($c=1/2$): $f'(1/2) = \frac{1}{2\sqrt{1/2}} = \frac{1}{\sqrt{2}} \neq \frac{1}{2}$. Option D ($c=3/2$): $f'(3/2) = \frac{1}{2\sqrt{3/2}} \neq \frac{1}{2}$.

Answer: At some instant during the trip, the vehicle's velocity was exactly 75 km/h.

Apply MVT to the position function: Let $s(t)$ be the position function. The average velocity is $\dfrac{s(2)-s(0)}{2-0} = \dfrac{150}{2} = 75$ km/h. By the MVT, there exists some $c \in (0,2)$ where $s'(c) = 75$ km/h. Why the correct answer works: The MVT guarantees that the instantaneous velocity equals the average velocity at least once during the trip. Why distractors fail: Option A claims constant velocity, which the MVT does not guarantee. Option C makes a claim about endpoint velocities, which is not part of the MVT conclusion. Option D claims a bound on maximum velocity, which the MVT does not establish.

Answer: No — while the MVT guarantees at least one $c$, solving $3c^2 = 1$ gives two values $c = \pm\dfrac{1}{\sqrt{3}}$, so uniqueness fails.

Verify the MVT applies: $f(x) = x^3$ is a polynomial, so it is continuous on $[-1, 1]$ and differentiable on $(-1, 1)$. The MVT applies. Find the average rate and solve: Average rate: $\dfrac{1 - (-1)}{1 - (-1)} = 1$. Setting $f'(c) = 3c^2 = 1$ gives $c^2 = \dfrac{1}{3}$, so $c = \pm\dfrac{1}{\sqrt{3}}$. Both values lie in $(-1, 1)$. Evaluate the student's claim: The student is wrong about uniqueness. The MVT guarantees existence of at least one $c$, not exactly one. Here there are two such values. Why distractors fail: Option A is wrong because the MVT is an existence theorem and does not guarantee uniqueness. Option C is wrong because polynomials are differentiable everywhere. Option D is wrong because even though $x^3$ is increasing, the equation $3c^2 = 1$ still has two solutions.

Answer: Show that $f'(x) = 5x^4 + 3 > 0$ for all $x$, so $f$ is strictly increasing and can cross zero at most once.

Identify the uniqueness strategy: A consequence of the MVT states that if $f'(x) > 0$ for all $x$, then $f$ is strictly increasing. A strictly increasing function can have at most one zero. Verify the derivative condition: $f'(x) = 5x^4 + 3$. Since $x^4 \geq 0$ for all real $x$, we have $f'(x) \geq 3 > 0$ everywhere. By the MVT consequence, $f$ is strictly increasing. Combine with IVT for the full result: IVT gives existence of a root (since $f(-1) = -3 < 0$ and $f(0) = 1 > 0$). The monotonicity argument from the MVT gives uniqueness. Why distractors fail: Option A focuses on where $f' = 0$, but $f'$ is never zero here, and even if it were, that alone doesn't determine the number of roots of $f$. Option C is wrong because odd-degree polynomials must have at least one real root, but can have more than one (e.g., $x^3 - x$). Option D is wrong because the sign change of $f''$ relates to concavity, not directly to the number of roots.