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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
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Tutti i libri di testoHass 15th EditionCh. 4 - Applications of DerivativesProblema 4.2.4
Capitolo 4, Problema 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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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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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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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) = x − 6√(x − 1)

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Identifying Extrema


In Exercises 15–18:


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


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


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26. Constructing cylinders Compare the answers to the following two construction problems.

a. A rectangular sheet of perimeter 36 cm and dimensions x cm by y cm is to be rolled into a cylinder as shown in part (a) of the figure. What values of x and y give the largest volume?

b. The same sheet is to be revolved about one of the sides of length y to sweep out the cylinder as shown in part (b) of the figure. What values of x and y give the largest volume?

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107. Marginal cost The accompanying graph shows the hypothetical cost c=f(x) of manufacturing x items. At approximately what production level does the marginal cost change from decreasing to increasing?

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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 = x¹¹ + x³ + x − 5

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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.

7. y=sin|x|, -2π≤x≤2π

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