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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoHass 15th EditionCh. 4 - Applications of DerivativesProblema 4.2.8
Capitolo 4, Problema 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]

Guida verificata passo dopo passo
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.

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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.
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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.
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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.
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Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.

85. y' = x^(-2/3) (x - 1)

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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/x² − x² − 1/3) dx

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35. Determine the dimensions of the rectangle of largest area that can be inscribed in the right triangle shown in the accompanying figure.

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Finding Critical Points


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


f(x) = x(4 − x)³

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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.

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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.

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