Mean Value Theorem - Calculus 1

Which of the following functions fails to satisfy the hypotheses of the Mean Value Theorem on the specified interval?
Answer: $h(x) = \frac{x^2 - 1}{x - 1}$ on $[0, 2]$ The Mean Value Theorem requires the function to be continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ . Let's check each function. (A) $f(x)=x^{2/3}$ is continuous on $[0, 1]$ and differentiable on $(0, 1)$ . (B) $g(x)=|x^2-4|$ is equal to $4-x^2$ on $[0, 1]$ , which is a polynomial and thus continuous and differentiable. (D) $k(x)=\ln(x-1)$ is continuous and differentiable for $x > 1$ , so it satisfies the hypotheses on $[2, 4]$ . (C) $h(x) = \frac{x^2 - 1}{x - 1}$ is not defined at $x=1$ . Since $1$ is in the interval $[0, 2]$ , the function is not continuous on the entire closed interval. Therefore, it fails to satisfy the hypotheses of the Mean Value Theorem.
Let ƒ be a differentiable function on [-2, 2] with ƒ(-2) = 1. If |ƒ'(x)| ≤ 3 for all x in (-2, 2), which of the following statements must be true?
Answer: ƒ(2) ≤ 13 By the Mean Value Theorem on [-2, 2], there exists a c in (-2, 2) such that ƒ'(c) = (ƒ(2) - ƒ(-2)) / (2 - (-2)) = (ƒ(2) - 1) / 4. We are given |ƒ'(x)| ≤ 3, which means -3 ≤ ƒ'(x) ≤ 3. So, -3 ≤ (ƒ(2) - 1) / 4 ≤ 3. Multiplying by 4 gives -12 ≤ ƒ(2) - 1 ≤ 12. Adding 1 to all parts gives -11 ≤ ƒ(2) ≤ 13. Therefore, it must be true that ƒ(2) ≤ 13.
Consider a function $f(x)$ which is continuous on $[a, b]$ and differentiable on $(a, b)$ . If $f(a) = f(b)$ , which of the following statements is a necessary consequence of the Mean Value Theorem?
Answer: There exists a point where $f'(x) = 0$ in $(a, b)$ . This scenario describes Rolle's Theorem, which is a special case of the Mean Value Theorem. Since $f(a) = f(b)$ , the average rate of change is $\frac{f(b) - f(a)}{b - a} = 0$ . By Rolle's Theorem, there exists at least one point $c$ in $(a, b)$ such that $f'(c) = 0$ . This means the function must have at least one critical point in the interval. The other choices are not necessary: the function need not be symmetric (consider $f(x) = x^2 - x$ on $[0,1]$ ), need not be constant (consider $f(x) = x(x-1)$ on $[0,1]$ ), and need not have a root in $(a,b)$ (consider $f(x) = x^2 - x + 2$ on $[0,1]$ ).
Let ƒ(x) = x³ - 6x² + 5x. On the interval [0, 4], the Mean Value Theorem guarantees the existence of at least one value c such that ƒ'(c) equals the average rate of change on the interval. What is the sum of all such values of c?
Answer: 4 First, find the average rate of change. ƒ(0) = 0. ƒ(4) = 4³ - 6(4)² + 5(4) = 64 - 96 + 20 = -12. The average rate of change is (ƒ(4) - ƒ(0)) / (4 - 0) = (-12 - 0) / 4 = -3. Next, find the derivative: ƒ'(x) = 3x² - 12x + 5. Set ƒ'(c) = -3: 3c² - 12c + 5 = -3, which simplifies to 3c² - 12c + 8 = 0. Using the quadratic formula, c = (12 ± √(144 - 4*3*8)) / 6 = (12 ± √48) / 6 = 2 ± (4√3)/6 = 2 ± (2√3)/3. Both c₁ = 2 - (2√3)/3 ≈ 0.845 and c₂ = 2 + (2√3)/3 ≈ 3.155 are in the interval (0, 4). The sum of these values is (2 - (2√3)/3) + (2 + (2√3)/3) = 4. Alternatively, by Vieta's formulas, the sum of the roots of 3c² - 12c + 8 = 0 is -(-12)/3 = 4.
Consider the function $f(x) = \arcsin(\frac{x}{2})$ . The Mean Value Theorem applied to $f(x)$ on the interval $[-1, 1]$ guarantees the existence of a number $c$ in $(-1, 1)$ . What is the value of $c^2$ ?
Answer: $4(1 - \frac{9}{\pi^2})$ First, find the average rate of change on $[-1, 1]$ . $f(1) = \arcsin(1/2) = \pi/6$ . $f(-1) = \arcsin(-1/2) = -\pi/6$ . The slope of the secant line is $\frac{\pi/6 - (-\pi/6)}{1 - (-1)} = \frac{2\pi/6}{2} = \frac{\pi}{6}$ . Next, find the derivative using the chain rule: $f'(x) = \frac{1}{\sqrt{1 - (x/2)^2}} \cdot \frac{1}{2} = \frac{1}{2\sqrt{1 - x^2/4}}$ . Set $f'(c) = \pi/6$ : $\frac{1}{2\sqrt{1 - c^2/4}} = \frac{\pi}{6}$ . This simplifies to $6 = 2\pi\sqrt{1 - c^2/4}$ , or $3/\pi = \sqrt{1 - c^2/4}$ . Squaring both sides gives $\frac{9}{\pi^2} = 1 - \frac{c^2}{4}$ . Rearranging terms, we get $\frac{c^2}{4} = 1 - \frac{9}{\pi^2}$ , so $c^2 = 4(1 - \frac{9}{\pi^2})$ .
Consider the function g(x) = x + cos(x) on the interval [0, π/2]. Let c be the value guaranteed by the Mean Value Theorem. Which of the following statements is true about c?
Answer: sin(c) = 2/π When you encounter a Mean Value Theorem problem, you need to apply the theorem systematically and then solve for the specific value it guarantees. The Mean Value Theorem states that for a continuous, differentiable function on $[a,b]$ , there exists some $c$ in $(a,b)$ where $f'(c) = \frac{f(b) - f(a)}{b - a}$ . For $g(x) = x + \cos(x)$ on $[0, \pi/2]$ , let's find this relationship. First, $g'(x) = 1 - \sin(x)$ . Next, we calculate the average rate of change: $\frac{g(\pi/2) - g(0)}{\pi/2 - 0} = \frac{(\pi/2 + 0) - (0 + 1)}{\pi/2} = \frac{\pi/2 - 1}{\pi/2} = 1 - \frac{2}{\pi}$ . Setting $g'(c) =$ average rate of change: $1 - \sin(c) = 1 - \frac{2}{\pi}$ . Solving: $-\sin(c) = -\frac{2}{\pi}$ , so $\sin(c) = \frac{2}{\pi}$ . Looking at the choices: A) $\sin(c) = \frac{2}{\pi}$ matches our result exactly. B) $c = \pi/6$ would give $\sin(c) = 1/2$ , which doesn't equal $2/\pi \approx 0.637$ . C) $c = \pi/4$ would give $\sin(c) = \sqrt{2}/2 \approx 0.707$ , also incorrect. D) $\cos(c) = 2/\pi$ confuses cosine with sine. Study tip: In Mean Value Theorem problems, always set up the equation $f'(c) = \frac{f(b) - f(a)}{b - a}$ first, then solve algebraically. Don't try to guess the value of $c$ directly—focus on what the theorem tells you about $f'(c)$ .
Let ƒ be a function that is continuous on [a, b] and differentiable on (a, b). If for every subinterval [x₁, x₂] ⊆ [a, b], the value c guaranteed by the Mean Value Theorem is always the midpoint of the subinterval, c = (x₁ + x₂) / 2, what type of function must ƒ(x) be?
Answer: A quadratic function The condition is (ƒ(x₂) - ƒ(x₁)) / (x₂ - x₁) = ƒ'((x₁ + x₂)/2). Let's test a general quadratic function, ƒ(x) = Ax² + Bx + C. The derivative is ƒ'(x) = 2Ax + B. The average rate of change is [A(x₂² - x₁²) + B(x₂ - x₁)] / (x₂ - x₁) = A(x₂ + x₁) + B. The instantaneous rate of change at the midpoint is ƒ'((x₁ + x₂)/2) = 2A((x₁ + x₂)/2) + B = A(x₁ + x₂) + B. The expressions match for any quadratic function. This property does not hold for cubic or higher-degree polynomials, and while it holds for linear functions (a subset of quadratics), 'quadratic' is the most general correct description.
Let $f$ be a twice-differentiable function. Suppose that $f'(1) = 4$ and $f'(5) = 12$ . Which of the following statements must be true?
Answer: There exists a number $c \in (1, 5)$ such that $f''(c) = 2$ . We can apply the Mean Value Theorem to the function $f'(x)$ on the interval $[1, 5]$ . Since $f$ is twice-differentiable, $f'(x)$ is continuous on $[1, 5]$ and differentiable on $(1, 5)$ . The MVT states there exists a $c \in (1, 5)$ such that $(f')'(c) = f''(c) = \frac{f'(5) - f'(1)}{5 - 1}$ . Plugging in the given values, we get $f''(c) = \frac{12 - 4}{4} = \frac{8}{4} = 2$ . Choice (A) is true by the Intermediate Value Theorem applied to $f'$ , but (B) is the direct consequence of the Mean Value Theorem. Choice (C) is not necessary; $f''(x)$ could be negative elsewhere in the interval. Choice (D) is incorrect; by the Fundamental Theorem of Calculus, the integral is $f(5) - f(1)$ , which is not necessarily 16.
Let ƒ be a differentiable function on [1, 5] such that ƒ(1) = 2. The derivative ƒ'(x) satisfies the inequality 1 ≤ ƒ'(x) ≤ 3 for all x in [1, 5]. Which of the following could be the value of ƒ(5)?
Answer: 10 By the Mean Value Theorem, there is a c in (1, 5) such that (ƒ(5) - ƒ(1)) / (5 - 1) = ƒ'(c). Substituting ƒ(1)=2, we get (ƒ(5) - 2) / 4 = ƒ'(c). Since 1 ≤ ƒ'(c) ≤ 3, we have 1 ≤ (ƒ(5) - 2) / 4 ≤ 3. Multiplying by 4 gives 4 ≤ ƒ(5) - 2 ≤ 12. Adding 2 to all parts gives 6 ≤ ƒ(5) ≤ 14. Of the given choices, only 10 falls within this interval [6, 14].
Let ƒ be a function that is continuous on [a, b] and differentiable on (a, b). If ƒ'(x) is never zero on (a, b), what is the maximum number of times the graph of ƒ can intersect the line y = ƒ(a)?
Answer: 1 The graph of ƒ intersects the line y = ƒ(a) at x = a. Suppose it intersects the line at another point, x = d, where a < d ≤ b. This would mean ƒ(a) = ƒ(d). By Rolle's Theorem (a special case of MVT), there must exist a c in (a, d) such that ƒ'(c) = 0. However, the problem states that ƒ'(x) is never zero on (a, b). This is a contradiction. Therefore, the graph cannot intersect the line y = ƒ(a) at any point other than x = a. The maximum number of intersections is 1.
For the function $f(x) = \frac{1}{x}$ on the interval $[2, 8]$ , what is the value of $c$ guaranteed by the Mean Value Theorem?
Answer: $4$ First, calculate the average rate of change of $f(x)$ on $[2, 8]$ . $f(8) = 1/8$ and $f(2) = 1/2$ . The slope of the secant line is $\frac{1/8 - 1/2}{8 - 2} = \frac{1/8 - 4/8}{6} = \frac{-3/8}{6} = -\frac{3}{48} = -\frac{1}{16}$ . Next, find the derivative of $f(x)$ : $f'(x) = -\frac{1}{x^2}$ . According to the Mean Value Theorem, we set $f'(c) = -\frac{1}{16}$ . This gives $-\frac{1}{c^2} = -\frac{1}{16}$ , which implies $c^2 = 16$ . The possible values for $c$ are $4$ and $-4$ . Since $c$ must be in the interval $(2, 8)$ , the only valid solution is $c=4$ .
Let ƒ be a twice-differentiable function. Suppose ƒ(1) = 3, ƒ(3) = 7, and ƒ(5) = 15. Which of the following statements must be true?
Answer: There exists a c in (1, 5) such that ƒ''(c) > 0. First, apply the Mean Value Theorem to ƒ on [1, 3]. There exists c₁ in (1, 3) such that ƒ'(c₁) = (ƒ(3) - ƒ(1)) / (3 - 1) = (7 - 3) / 2 = 2. Next, apply the MVT to ƒ on [3, 5]. There exists c₂ in (3, 5) such that ƒ'(c₂) = (ƒ(5) - ƒ(3)) / (5 - 3) = (15 - 7) / 2 = 4. Now, apply the MVT to the function ƒ' on the interval [c₁, c₂]. Since ƒ is twice-differentiable, ƒ' is differentiable. There exists d in (c₁, c₂) such that ƒ''(d) = (ƒ'(c₂) - ƒ'(c₁)) / (c₂ - c₁). Substituting the values, ƒ''(d) = (4 - 2) / (c₂ - c₁) = 2 / (c₂ - c₁). Since c₂ > c₁, the denominator is positive, so ƒ''(d) > 0. As (c₁, c₂) is a subinterval of (1, 5), d is in (1, 5).
Let ƒ(x) = 1/x on the interval [1, k] where k > 1. If the value c guaranteed by the Mean Value Theorem is c = √3, what is the value of k?
Answer: 3 First, calculate the average rate of change: (ƒ(k) - ƒ(1)) / (k - 1) = (1/k - 1) / (k - 1) = ((1-k)/k) / (k-1) = -1/k. Next, calculate the derivative: ƒ'(x) = -1/x². According to the Mean Value Theorem, ƒ'(c) equals the average rate of change. We are given c = √3. So, ƒ'(√3) = -1/(√3)² = -1/3. Setting the two expressions equal: -1/k = -1/3. Therefore, k = 3.
Let $f(x)$ be a differentiable function on the interval $[1, 5]$ . If the Mean Value Theorem guarantees a value $c=3$ such that $f'(3) = 10$ , what is the value of $f(5) - f(1)$ ?
Answer: $40$ The Mean Value Theorem states that there is a number $c$ in $(a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$ . In this case, $a=1$ , $b=5$ , and we are given $c=3$ and $f'(3)=10$ . Plugging these values into the formula gives $10 = \frac{f(5) - f(1)}{5 - 1}$ . This simplifies to $10 = \frac{f(5) - f(1)}{4}$ . Multiplying both sides by 4 gives $f(5) - f(1) = 40$ .
A particle's position is given by s(t), a differentiable function of time t. At t = 1 hour, the particle is at position s(1) = 10 km. At t = 5 hours, it is at s(5) = 90 km. Which of the following statements must be true as a consequence of the Mean Value Theorem?
Answer: There is at least one time t in (1, 5) at which the particle's instantaneous velocity is exactly 20 km/hr. The average velocity of the particle over the interval [1, 5] is (s(5) - s(1)) / (5 - 1) = (90 - 10) / 4 = 80 / 4 = 20 km/hr. The Mean Value Theorem states that since s(t) is differentiable, there must be at least one time c in the open interval (1, 5) where the instantaneous velocity, s'(c), is equal to the average velocity. Therefore, s'(c) = 20 km/hr for some c in (1, 5). The theorem does not specify the value of c, only its existence.
The Mean Value Theorem does not apply to ƒ(x) = |x² - 4| on the interval [-3, 3]. Which hypothesis of the theorem is not satisfied?
Answer: ƒ(x) is not differentiable on (-3, 3). The Mean Value Theorem is a fundamental result in calculus that requires two key conditions: the function must be continuous on the closed interval and differentiable on the open interval. When you encounter questions about MVT applications, always check both conditions systematically. For $f(x) = |x² - 4|$ on $[-3, 3]$ , you need to analyze where the absolute value creates problems. First, determine where $x² - 4 = 0$ , which occurs at $x = ±2$ . At these points, the expression inside the absolute value changes sign, so $f(x) = |x² - 4|$ becomes: - $f(x) = x² - 4$ when $x² - 4 ≥ 0$ (outside $[-2, 2]$ ) - $f(x) = -(x² - 4) = 4 - x²$ when $x² - 4 < 0$ (inside $(-2, 2)$ ) The function is continuous everywhere since absolute value functions are continuous. However, at $x = ±2$ , the derivative changes abruptly from $2x$ to $-2x$ , creating sharp corners where the function isn't differentiable. Choice A is correct because differentiability fails at $x = ±2$ within the open interval $(-3, 3)$ . Choice B is wrong because $f(x)$ is clearly defined everywhere on $[-3, 3]$ . Choice C is incorrect since absolute value functions maintain continuity. Choice D is irrelevant because MVT doesn't require the endpoint values to be equal—that's Rolle's Theorem. Study tip: When checking MVT applicability, always look for sharp corners or cusps in absolute value functions by finding where the inside expression equals zero.
Let ƒ be a function that is differentiable on (0, ∞) and continuous on [0, ∞). If ƒ(0) = -3 and |ƒ'(x)| ≤ 1 for all x > 0, what is the largest possible value of ƒ(4)?
Answer: 1 By the Mean Value Theorem on [0, 4], there exists a c in (0, 4) such that ƒ'(c) = (ƒ(4) - ƒ(0)) / (4 - 0). We are given ƒ(0) = -3 and |ƒ'(x)| ≤ 1, which implies -1 ≤ ƒ'(x) ≤ 1. So, -1 ≤ (ƒ(4) - (-3)) / 4 ≤ 1. This means -1 ≤ (ƒ(4) + 3) / 4 ≤ 1. Multiplying by 4 gives -4 ≤ ƒ(4) + 3 ≤ 4. Subtracting 3 from all parts gives -7 ≤ ƒ(4) ≤ 1. The largest possible value of ƒ(4) is 1.
Let ƒ be a differentiable function such that ƒ(0) = 0 and ƒ'(x) ≤ 1 for all x ≥ 0. Which of the following values is not a possible value for ƒ(2)?
Answer: 3 By the Mean Value Theorem on the interval [0, 2], there is a c in (0, 2) such that ƒ'(c) = (ƒ(2) - ƒ(0)) / (2 - 0). Substituting the given information, we have ƒ'(c) = (ƒ(2) - 0) / 2 = ƒ(2) / 2. We are also given that ƒ'(x) ≤ 1 for all x ≥ 0, which includes c. So, ƒ'(c) ≤ 1. This implies ƒ(2) / 2 ≤ 1, which means ƒ(2) ≤ 2. Of the given choices, only 3 is greater than 2, so it is not a possible value for ƒ(2).
The Mean Value Theorem requires a function ƒ to be continuous on [a, b] and differentiable on (a, b). Which of the following function-interval pairs demonstrates that the differentiability condition is necessary for the theorem's conclusion to hold?
Answer: ƒ(x) = |x| on [-1, 1] We need a function that is continuous on the closed interval but not differentiable on the open interval, and for which the conclusion of the MVT fails. For ƒ(x) = |x| on [-1, 1], the function is continuous. However, it is not differentiable at x=0. The average rate of change is (ƒ(1) - ƒ(-1)) / (1 - (-1)) = (1 - 1) / 2 = 0. The derivative is ƒ'(x) = -1 for x < 0 and ƒ'(x) = 1 for x > 0. There is no c in (-1, 1) where ƒ'(c) = 0. Thus, the conclusion fails, demonstrating the necessity of the differentiability hypothesis. Choice C fails the continuity condition. Choices A and D satisfy all hypotheses.
Let ƒ(x) = ln(x). On the interval [1, e²], the Mean Value Theorem guarantees the existence of a point c where the tangent line to the graph of ƒ is parallel to the secant line through the interval's endpoints. What is the value of c?
Answer: (e² - 1) / 2 The Mean Value Theorem states that for a continuous, differentiable function on a closed interval, there exists at least one point where the instantaneous rate of change (derivative) equals the average rate of change over the interval. To find this point c, you need to set up the equation $f'(c) = \frac{f(b) - f(a)}{b - a}$ . For $f(x) = \ln(x)$ on $[1, e^2]$ , first calculate the average rate of change: $\frac{\ln(e^2) - \ln(1)}{e^2 - 1} = \frac{2 - 0}{e^2 - 1} = \frac{2}{e^2 - 1}$ . Since $f'(x) = \frac{1}{x}$ , you need to solve $\frac{1}{c} = \frac{2}{e^2 - 1}$ . Cross-multiplying gives $e^2 - 1 = 2c$ , so $c = \frac{e^2 - 1}{2}$ . Choice A is correct: $\frac{e^2 - 1}{2}$ . Choice B ( $e^2 - 1$ ) is twice the correct answer—this would result from forgetting to divide by 2 when solving for c. Choice C ( $e$ ) might seem tempting as the midpoint of the interval in some sense, but it doesn't satisfy the MVT equation. Choice D ( $\frac{2}{e^2 - 1}$ ) is actually the slope value itself, not the x-coordinate where this slope occurs. Remember: the Mean Value Theorem asks where the derivative equals the average rate of change. Always calculate the average rate first, then set it equal to the derivative and solve for the unknown point.

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