Skip to main content
Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 4 - Applications of DerivativesProblem 4.2.6
Chapter 4, Problem 4.2.6

Checking the Mean Value Theorem


Find the value or values of c that satisfy the equation (f(b) − f(a)) / (b − a) = f′(c) in the conclusion of the Mean Value Theorem for the functions and intervals in Exercises 1–6.


g(x) = {x³, −2 ≤ x ≤ 0
x², 0 < x ≤ 2

Verified step by step guidance
1
First, understand the Mean Value Theorem (MVT). It states that for a function g(x) that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), there exists at least one c in (a, b) such that (g(b) - g(a)) / (b - a) = g'(c).
Identify the intervals and the function pieces: g(x) = x³ for -2 ≤ x ≤ 0 and g(x) = x² for 0 < x ≤ 2. The function is continuous on [-2, 2] but we need to check differentiability at x = 0.
Check differentiability at x = 0 by finding the derivatives of each piece: For x³, the derivative is g'(x) = 3x², and for x², the derivative is g'(x) = 2x. Evaluate the left-hand and right-hand derivatives at x = 0 to ensure they are equal.
Calculate g(-2) and g(2) to apply the MVT: g(-2) = (-2)³ = -8 and g(2) = (2)² = 4. Use these to find the average rate of change: (g(2) - g(-2)) / (2 - (-2)) = (4 - (-8)) / 4.
Set the average rate of change equal to the derivative g'(c) and solve for c in each interval: For -2 < c < 0, use g'(c) = 3c², and for 0 < c < 2, use g'(c) = 2c. Solve these equations to find the value(s) of c that satisfy the MVT.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
6m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Mean Value Theorem

The Mean Value Theorem (MVT) states that for a continuous function on a closed interval [a, b] that is differentiable on the open interval (a, b), there exists at least one point c in (a, b) such that the derivative at c equals the average rate of change over [a, b]. This is expressed as f'(c) = (f(b) - f(a)) / (b - a).
Recommended video:
06:11
Fundamental Theorem of Calculus Part 1

Piecewise Functions

A piecewise function is defined by different expressions over different intervals. In this problem, g(x) is defined as x³ for -2 ≤ x ≤ 0 and x² for 0 < x ≤ 2. Understanding how to evaluate and differentiate each piece within its respective interval is crucial for applying the Mean Value Theorem.
Recommended video:
05:36
Piecewise Functions

Differentiability and Continuity

For the Mean Value Theorem to apply, the function must be continuous on the closed interval and differentiable on the open interval. This requires checking that g(x) is continuous at x = 0, where the pieces meet, and ensuring differentiability across the entire interval, except possibly at the endpoints.
Recommended video:
05:34
Intro to Continuity
Related Practice
Textbook Question

117. Suppose that the second derivative of the function y = f(x) isy" =(x+1)(x-2).

For what x-values does the graph of f have an inflection point?

222
views
Textbook Question

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


dv/dt = (1/2)sec t tan t, v(0) = 1

13
views
Textbook Question

Identifying Extrema


In Exercises 19–40:


a. Find the open intervals on which the function is increasing and those on which it is decreasing.


b. Identify the function’s local extreme values, if any, saying where they occur.


f(x) = x³ / (3x² + 1)

159
views
Textbook Question

Finding Critical Points


In Exercises 41–50, determine all critical points and all domain endpoints for each function.


y = x² + 2/x

159
views
Textbook Question

"Roots (Zeros) Show that the functions in Exercises 19–26 have exactly one zero

206
views
Textbook Question

Finding Critical Points


In Exercises 41–50, determine all critical points and all domain endpoints for each function.


y = x² − 32√x

200
views