Textbook Question
99. In Exercises 99 and 100, the graph of f' is given. Determine x-values corresponding to inflection points for the graph of f.
Ch. 4 - Applications of Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
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 guidance1
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:
4mWas 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:
04:50Critical 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:
4:26Evaluating Composed Functions
Related Practice
Textbook Question
Initial Value Problems
Solve the initial value problems in Exercises 71–90.
dr/dθ = −π sin (πθ), r(0) = 0
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.
Textbook Question
Solve the initial value problems in Exercises 71–90.
y⁽⁴⁾ = −sin t + cos t;
y′′′(0) =7, y′′(0) = y′(0) = −1, y(0) = 0
Textbook Question
Business and Economics
60. Production level Prove that the production level (if any) at which average cost is smallest is a level at which the average cost equals marginal cost.
Textbook Question
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.
∫x⁻¹ᐟ³ dx