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 4.2.8
Chapter 4, Problem 4.2.8

Checking the Mean Value Theorem


Which of the functions in Exercises 7–12 satisfy the hypotheses of the Mean Value Theorem on the given interval, and which do not? Give reasons for your answers.


f(x) = x⁴ᐟ⁵, [0, 1]

Verified step by step guidance
1
Step 1: Recall the Mean Value Theorem (MVT), which states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one c in (a, b) such that f'(c) = (f(b) - f(a)) / (b - a).
Step 2: Check the continuity of f(x) = x^(4/5) on the interval [0, 1]. Since x^(4/5) is a root function, it is continuous on [0, 1] because the domain of x^(4/5) is all non-negative x.
Step 3: Check the differentiability of f(x) = x^(4/5) on the interval (0, 1). The derivative f'(x) = (4/5)x^(-1/5) is defined for all x > 0, so f is differentiable on (0, 1).
Step 4: Consider the endpoint x = 0. The derivative f'(x) = (4/5)x^(-1/5) is not defined at x = 0, which means f is not differentiable at the endpoint x = 0.
Step 5: Conclude that the function f(x) = x^(4/5) does not satisfy the hypotheses of the Mean Value Theorem on the interval [0, 1] because it is not differentiable at x = 0.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
5m
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 function f that is continuous on a 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) equals the average rate of change over [a, b]. This theorem helps in understanding the behavior of functions and is crucial for verifying if a function meets its conditions.
Recommended video:
06:11
Fundamental Theorem of Calculus Part 1

Continuity

Continuity of a function on an interval means that the function has no breaks, jumps, or holes in that interval. For the Mean Value Theorem to apply, the function must be continuous on the closed interval [a, b]. This ensures that the function behaves predictably and smoothly across the entire interval, which is necessary for finding a point where the instantaneous rate of change matches the average rate of change.
Recommended video:
05:34
Intro to Continuity

Differentiability

Differentiability refers to the existence of a derivative at each point in an interval. For the Mean Value Theorem, the function must be differentiable on the open interval (a, b). Differentiability implies continuity, but not vice versa, and ensures that the function has a well-defined tangent at every point in the interval, which is essential for applying the theorem.
Recommended video:
05:53
Finding Differentials
Related Practice
Textbook Question

Theory and Examples


In Exercises 53 and 54, show that the function has neither an absolute minimum nor an absolute maximum on its natural domain.


y = 3x + tan x

Textbook Question

41. Among all triangles in the first quadrant formed by the x-axis, the y-axis, and tangent lines to the graph of y=3x-x^2, what is the smallest possible area?

Textbook Question

22. A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only half as much light per unit area as clear glass does. The total perimeter is fixed. Find the proportions of the window that will admit the most light. Neglect the thickness of the frame.

" style="" width="180">

Textbook Question

Find values of a and b such that the function


ƒ(𝓍) = (a𝓍 + b) / 𝓍² ―1)


has a local extreme value of 1 at 𝓍 = 3. Is this extreme value a local maximum or a local minimum? Give reasons for your answer.

Textbook Question

Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the open intervals on which the functions are differentiable and the graphs are concave up and concave down.

4. y=9/14x^(1/3)(x^2-7)

Textbook Question

Finding Indefinite Integrals


In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.


∫(1 + tan²θ)dθ (Hint:1 + tan²θ = sec²θ)