Textbook Question
A cone is formed from a circular piece of material of radius 1 meter by removing a section of angle θ and then joining the two straight edges. Determine the largest possible volume for the cone.
Ch. 4 - Applications of Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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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 4.4.2
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.
2. y=x^4/4-2x^2+4
Verified step by step guidance1
To find the inflection points, local maxima, and minima, start by finding the first derivative of the function y = \(\frac{x^4}{4}\) - 2x^2 + 4. The first derivative, y', will help identify critical points where the slope is zero or undefined.
Calculate the first derivative: y' = \(\frac{d}{dx}\)(\(\frac{x^4}{4}\) - 2x^2 + 4). This simplifies to y' = x^3 - 4x.
Set the first derivative equal to zero to find critical points: x^3 - 4x = 0. Factor the equation: x(x^2 - 4) = 0, which gives x = 0, x = 2, and x = -2.
To determine if these critical points are local maxima, minima, or points of inflection, calculate the second derivative: y'' = \(\frac{d}{dx}\)(x^3 - 4x). This simplifies to y'' = 3x^2 - 4.
Evaluate the second derivative at the critical points: y''(0), y''(2), and y''(-2). If y'' > 0, the function is concave up (local minima); if y'' < 0, the function is concave down (local maxima); if y'' = 0, it may be an inflection point. Analyze the sign changes in y'' to identify intervals of concavity.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Inflection Points
Inflection points are points on a curve where the concavity changes, meaning the curve transitions from being concave up to concave down or vice versa. To find inflection points, one must analyze the second derivative of the function. If the second derivative changes sign at a point, that point is an inflection point.
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Critical Points
Local Maxima and Minima
Local maxima and minima are points where a function reaches a peak or a trough within a certain interval. These points can be identified using the first derivative test, where critical points (where the first derivative is zero or undefined) are evaluated to determine if they correspond to a maximum, minimum, or neither based on the sign of the derivative before and after the point.
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07:09The First Derivative Test: Finding Local Extrema
Concavity and Differentiability
Concavity refers to the direction of the curvature of a function, determined by the sign of the second derivative. A function is concave up when its second derivative is positive and concave down when it is negative. Differentiability indicates that a function has a derivative at all points in an interval, which is essential for determining local extrema and concavity.
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05:59Determining Concavity Given a Function
Related Practice
Textbook Question
Finding Extrema from Graphs
In Exercises 7–10, find the absolute extreme values and where they occur.
Textbook Question
Identifying Extrema
In Exercises 61 and 62, the graph of f' is given. Assume that f is continuous, and determine the x-values corresponding to local minima and local maxima.
Textbook Question
Identifying Extrema
In Exercises 61 and 62, the graph of f' is given. Assume that f is continuous, and determine the x-values corresponding to local minima and local maxima.
Textbook Question
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.
77. y' = cot(θ/2), for 0 < θ < 2π
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.
∫(1 + cos 4t)/2 dt