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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 4.2.4
Chapter 4, Problem 4.2.4

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.


f(x) =√(x − 1), [1, 3]

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1
First, ensure that the function f(x) = √(x - 1) is continuous on the closed interval [1, 3] and differentiable on the open interval (1, 3). Since the square root function is continuous and differentiable wherever its argument is positive, f(x) meets these conditions on the given interval.
Calculate f(a) and f(b) where a = 1 and b = 3. This involves evaluating the function at the endpoints of the interval: f(1) = √(1 - 1) = 0 and f(3) = √(3 - 1) = √2.
Apply the Mean Value Theorem formula: (f(b) - f(a)) / (b - a) = f'(c). Substitute the values: (√2 - 0) / (3 - 1) = f'(c). Simplify the left side to get √2 / 2.
Find the derivative f'(x) of the function f(x) = √(x - 1). Using the chain rule, f'(x) = 1/(2√(x - 1)).
Set the derivative equal to the simplified expression from the Mean Value Theorem: 1/(2√(c - 1)) = √2 / 2. Solve this equation for c to find the value(s) of c that satisfy the Mean Value Theorem.

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Key Concepts

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

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) equals the average rate of change over [a, b]. This theorem connects the derivative of a function to its overall behavior on an interval.
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Fundamental Theorem of Calculus Part 1

Continuity and Differentiability

For the Mean Value Theorem to apply, the function must be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Continuity ensures no breaks or jumps in the function, while differentiability ensures the function has a defined slope at every point within the interval. These conditions are crucial for finding the point c where the theorem holds.
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Intro to Continuity

Derivative Calculation

Calculating the derivative of the function f(x) = √(x − 1) is essential to apply the Mean Value Theorem. The derivative, f'(x), represents the instantaneous rate of change of the function. For f(x) = √(x − 1), using the chain rule, f'(x) = 1/(2√(x − 1)). This derivative helps find the specific value of c that satisfies the theorem's conclusion.
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Related Practice
Textbook Question

Finding Extrema from Graphs


In Exercises 7–10, find the absolute extreme values and where they occur.


Textbook Question

Motion with constant acceleration The standard equation for the position s of a body moving with a constant acceleration a along a coordinate line is s = (a/2)t² + v₀t + s₀, where v₀ and s₀ are the body’s velocity and position at time t = 0. Derive this equation by solving the initial value problem

Differential equation: d²s/dt² = a

Initial conditions: ds/dt = v₀ and s = s₀ when t=0.

Textbook Question

Applications


A marathoner ran the 26.2-mi New York City Marathon in 2.2 hours. Show that at least twice the marathoner was running at exactly 11 mph, assuming the initial and final speeds are zero.

Textbook Question

Solve the initial value problems in Exercises 71–90.

y⁽⁴⁾ = −sin t + cos t;

y′′′(0) =7, y′′(0) = y′(0) = −1, y(0) = 0

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.


∫x⁻¹ᐟ³ dx

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.


∫csc θ/(csc θ − sin θ) dθ