Skip to main content
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 4.1.39
Chapter 4, Problem 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

Verified step by step guidance
1
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.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
4m
Was this helpful?

Key Concepts

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

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.
Recommended video:
05:58
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.
Recommended video:
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.
Recommended video:
4:26
Evaluating Composed Functions
Related Practice
Textbook Question

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

215
views
Textbook Question

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).

215
views
Textbook Question

Finding Extrema from Graphs


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


321
views
Textbook Question

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


dr/dθ = −π sin (πθ), r(0) = 0

25
views
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.

52
views
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.

213
views