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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 4.1.53
Capitolo 4, Problema 4.1.53

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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First, identify the natural domain of the function y = x¹¹ + x³ + x − 5. Since this is a polynomial function, its domain is all real numbers, ℝ.
Next, consider the behavior of the function as x approaches positive and negative infinity. For large values of x, the term x¹¹ will dominate the behavior of the function because it has the highest degree.
Analyze the leading term x¹¹: As x approaches positive infinity, x¹¹ also approaches positive infinity, and as x approaches negative infinity, x¹¹ approaches negative infinity. This indicates that the function does not have a bound in either direction.
To further investigate, find the derivative of the function, y' = 11x¹⁰ + 3x² + 1, to determine critical points where the function might have local extrema. Set y' = 0 and solve for x to find these points.
Evaluate the second derivative, y'' = 110x⁹ + 6x, to determine the concavity at the critical points. This will help confirm whether these points are local minima or maxima. However, since the function is unbounded, these local extrema cannot be absolute extrema.

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Natural Domain

The natural domain of a function is the set of all real numbers for which the function is defined. For polynomial functions like y = x¹¹ + x³ + x − 5, the natural domain is all real numbers, as polynomials are defined for every real number without restrictions such as division by zero or square roots of negative numbers.
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Absolute Extrema

Absolute extrema refer to the highest or lowest points (maximum or minimum) of a function over its entire domain. A function has an absolute maximum at a point if its value there is greater than or equal to its value at any other point in the domain, and an absolute minimum if its value is less than or equal to any other point. For polynomials of odd degree, like y = x¹¹ + x³ + x − 5, the function tends to infinity as x approaches positive or negative infinity, often resulting in no absolute extrema.
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Finding Extrema Graphically

Behavior of Polynomial Functions

The behavior of polynomial functions, especially those of odd degree, is crucial in determining the presence of absolute extrema. Odd-degree polynomials, such as y = x¹¹ + x³ + x − 5, have end behaviors where one end goes to positive infinity and the other to negative infinity. This characteristic implies that such functions do not have absolute maxima or minima, as they do not level off at any finite value across their domain.
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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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10. Catching rainwater A 1125 ft^3 open-top rectangular tank with a square base x ft on a side and y ft deep is to be built with its top flush with the ground to catch runoff water. The costs associated with the tank involve not only the material from which the tank is made but also an excavation charge proportional to the product xy.

a. If the total cost is c=5(x^2+4xy) + 10xy, what values of x and y will minimize it?

b. Give a possible scenario for the cost function in part (a).

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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.

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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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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.

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