Textbook Question
Applications
Liftoff from Earth A rocket lifts off the surface of Earth with a constant acceleration of 20 m/sec². How fast will the rocket be going 1 min later?
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 Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 4, Problem 4.2
Finding Extreme Values
In Exercises 1–10, find the extreme values (absolute and local) of the function over its natural domain, and where they occur.
y = 𝓍³ ― 2𝓍 + 4
Verified step by step guidance1
First, find the derivative of the function \( y = x^3 - 2x + 4 \) to determine the critical points. The derivative is \( y' = 3x^2 - 2 \).
Set the derivative equal to zero to find the critical points: \( 3x^2 - 2 = 0 \). Solve for \( x \) to find the critical points.
Solve the equation \( 3x^2 - 2 = 0 \) by isolating \( x^2 \): \( x^2 = \frac{2}{3} \). Then, take the square root of both sides to find \( x = \pm \sqrt{\frac{2}{3}} \).
Evaluate the second derivative \( y'' = 6x \) to determine the concavity at each critical point. Substitute each critical point into the second derivative to determine if it is a local minimum or maximum.
Finally, evaluate the original function \( y = x^3 - 2x + 4 \) at the critical points and endpoints of the domain (if any) to find the absolute extreme values. Compare these values to determine the absolute maximum and minimum.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Critical Points
Critical points of a function occur where its derivative is zero or undefined. These points are potential locations for local extrema. To find them, compute the derivative of the function and solve for values of x where the derivative equals zero or does not exist. For the function y = x³ - 2x + 4, find the derivative and solve for x to identify critical points.
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04:50
Critical Points
First Derivative Test
The First Derivative Test helps determine whether a critical point is a local maximum, minimum, or neither. By analyzing the sign of the derivative before and after the critical point, one can infer the behavior of the function. If the derivative changes from positive to negative, the point is a local maximum; if it changes from negative to positive, it's a local minimum.
Recommended video:
07:09The First Derivative Test: Finding Local Extrema
Second Derivative Test
The Second Derivative Test provides another method to classify critical points. If the second derivative at a critical point is positive, the function has a local minimum there; if negative, a local maximum. If the second derivative is zero, the test is inconclusive. For y = x³ - 2x + 4, compute the second derivative to apply this test at the critical points.
Recommended video:
06:02The Second Derivative Test: Finding Local Extrema
Related Practice
Textbook Question
In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.
53. y = x * √(8 - x²)
Textbook Question
Finding Critical Points
In Exercises 41–50, determine all critical points and all domain endpoints for each function.
y = x² + 2/x
Textbook Question
In Exercises 1–10, find the extreme values (absolute and local) of the function over its natural domain, and where they occur.
______
y = √𝓍² ― 1
Textbook Question
54. Fermat’s principle in optics Light from a source A is reflected by a plane mirror to a receiver at point B, as shown in the accompanying figure. Show that for the light to obey Fermat’s principle, the angle of incidence must equal the angle of reflection, both measured from the line normal to the reflecting surface. (This result can also be derived without calculus. There is a purely geometric argument, which you may prefer.)
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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 + ³√x) dx