5.1 The Mean Value Theorem Homework Answer Key

5.1 the meanvalue theorem homework answer key – Understanding how to apply the Mean Value Theorem (MVT) is a cornerstone of introductory calculus, and having a clear answer key can turn confusing textbook problems into confidence‑boosting victories. This guide walks you through the theorem’s statement, the step‑by‑step process for solving typical homework questions, and detailed solutions to common exercises. By the end, you’ll not only know the correct answers but also the reasoning that makes each step logical and memorable.

Understanding the Mean Value Theorem

The Mean Value Theorem states that for a function f that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), there exists at least one point c in (a, b) such that

[ f'(c)=\frac{f(b)-f(a)}{b-a}. ]

In plain English, the theorem guarantees that somewhere between a and b the instantaneous rate of change (the derivative) matches the average rate of change over the entire interval. This idea appears repeatedly in homework sections, especially in sections like 5.1, where instructors test whether students can locate the required c and verify the conditions.

Key Conditions

1. Continuity on [a, b] – The function must not have any breaks, jumps, or holes on the closed interval.
2. Differentiability on (a, b) – The function must have a defined derivative at every interior point; corners or cusps disqualify the function.

If either condition fails, the theorem does not apply, and the homework problem may ask you to explain why no such c exists.

How to Approach MVT Homework Problems

When tackling a problem labeled 5.1, follow this systematic checklist:

1. Identify the interval [a, b] given in the question.
2. Check continuity on [a, b]. If the function is a polynomial, rational function (with denominator non‑zero), or trigonometric function over the interval, it is automatically continuous.
3. Check differentiability on (a, b). Again, most elementary functions meet this criterion unless they involve absolute values or piecewise definitions at the endpoints.
4. Compute the average rate of change: (\displaystyle \frac{f(b)-f(a)}{b-a}).
5. Find the derivative f'(x).
6. Solve the equation (f'(c)=\frac{f(b)-f(a)}{b-a}) for c within (a, b).
7. Verify that the solution(s) lie strictly inside the interval; if not, revisit steps 2–3.

Using this workflow ensures you never miss a hidden restriction and that your answer key will be both correct and defensible.

Sample Homework Problems and Their Solutions

Below are three representative problems that often appear in a 5.1 the mean value theorem homework answer key. Each includes a brief statement, the logical steps, and the final answer.

Problem 1: Polynomial Function

Given (f(x)=x^{3}-3x+2) on the interval [0, 2].
Find the value(s) of c that satisfy the Mean Value Theorem.

Solution

1. Continuity & Differentiability – f is a polynomial, so it is continuous on [0, 2] and differentiable on (0, 2).
2. Average rate of change:

[\frac{f(2)-f(0)}{2-0}= \frac{(8-6+2)-(0-0+2)}{2}= \frac{4-2}{2}=1. ]

3. Derivative: (f'(x)=3x^{2}-3).
4. Set derivative equal to 1:

[ 3c^{2}-3 = 1 ;\Longrightarrow; 3c^{2}=4 ;\Longrightarrow; c^{2}= \frac{4}{3} ;\Longrightarrow; c= \pm \frac{2}{\sqrt{3}}. ]

5. Select the value inside (0, 2): Only (c = \frac{2}{\sqrt{3}} \approx 1.155) lies in the interval.

Answer Key: (c = \dfrac{2}{\sqrt{3}}).

Problem 2: Trigonometric Function

Given (g(x)=\sin x) on [0, \pi].
Determine the point(s) c where the MVT holds.

Solution

1. g is continuous on [0, \pi] and differentiable on (0, \pi). 2. Average rate of change: [ \frac{\sin(\pi)-\sin(0)}{\pi-0}= \frac{0-0}{\pi}=0. ]

2. Derivative: (g'(x)=\cos x).

3. Solve (\cos c = 0) for c in (0, \pi). The cosine function equals zero at (c = \frac{\pi}{2}).

Answer Key: (c = \dfrac{\pi}{2}).

Problem 3: Rational Function with a Hole

Given (h(x)=\dfrac{x^{2}-4}{x-2}) on [1, 3].
Find the c that satisfies the MVT.

Solution

1. Simplify the function first:

[ h(x)=\dfrac{(x-2)(x+2)}{x-2}=x+2 \quad \text{for } x\neq 2. ]

The original expression is undefined at x = 2, but the simplified form shows the function behaves like a straight line everywhere else.

2. Continuity – The simplified function x+2 is continuous on [1, 3], but the original h has a removable discontinuity at x = 2. Since the interval includes the point of discontinuity, we must treat the function as not continuous on the closed interval.

3. Because the continuity condition fails, the Mean Value Theorem does not apply, and there is no c that satisfies the theorem for the original h.

Answer Key: No value of c exists; the theorem’s hypotheses are violated.

Common Mistakes and How to Avoid Them

• Skipping the continuity check – Even if a function looks “nice,” piecewise definitions can hide jumps. Always verify the interval endpoints.
• Including endpoint values for c – The theorem requires c to be strictly inside the interval, not at a or b.
• Algebraic errors when solving for c – Double‑check each algebraic manipulation, especially when dealing with fractions or square roots.
• **Misident