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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
4장, 문제 4.1.39

Absolute Extrema on Finite Closed Intervals


In Exercises 37–40, find the function’s absolute maximum and minimum values and say where they occur.


g(θ) = θ³ᐟ⁵, −32 ≤ θ ≤ 1

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First, understand that finding absolute extrema on a closed interval involves evaluating the function at critical points and endpoints of the interval. The function given is g(θ) = θ³ᐟ⁵, and the interval is [-32, 1].
To find critical points, calculate the derivative of the function g(θ). The derivative, g'(θ), can be found using the power rule for derivatives. For g(θ) = θ³ᐟ⁵, the derivative is g'(θ) = (3/5)θ⁻²ᐟ⁵.
Set the derivative g'(θ) equal to zero to find critical points. Solve the equation (3/5)θ⁻²ᐟ⁵ = 0. Since the derivative is never zero for any real θ, there are no critical points from setting the derivative to zero.
Next, evaluate the function g(θ) at the endpoints of the interval. Calculate g(-32) and g(1) to determine the values of the function at these points.
Compare the values of g(θ) at the endpoints to determine the absolute maximum and minimum values of the function on the interval [-32, 1]. The largest value is the absolute maximum, and the smallest value is the absolute minimum.

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Absolute Extrema

Absolute extrema refer to the highest and lowest values a function can achieve on a given interval. To find these values, evaluate the function at critical points and endpoints of the interval. The absolute maximum is the largest value, and the absolute minimum is the smallest value within the specified range.
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Finding Extrema Graphically

Critical Points

Critical points are values of the variable where the derivative of the function is zero or undefined. These points are potential candidates for local maxima or minima. In the context of finding absolute extrema, critical points within the interval are evaluated alongside the endpoints to determine the function's extreme values.
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04:50
Critical Points

Evaluating Function at Endpoints

Evaluating the function at the endpoints of the interval is crucial when determining absolute extrema. The endpoints are the boundary values of the interval, and the function's value at these points must be compared with values at critical points to identify the absolute maximum and minimum values on the interval.
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Evaluating Composed Functions
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교과서 질문

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.

8. y = 2cosx - √2x, -π≤x≤3π/2

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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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교과서 질문

Finding Extrema from Graphs


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


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Initial Value Problems


Solve the initial value problems in Exercises 71–90.


dr/dθ = −π sin (πθ), r(0) = 0

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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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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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