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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 4 - Applications of the DerivativeProblem 4.1.45
Chapter 4, Problem 4.1.45

Absolute maxima and minima Determine the location and value of the absolute extreme values of ƒ on the given interval, if they exist.


ƒ(x) = x³ - 3x² on [-1, 3]

Verified step by step guidance
1
First, understand that absolute extrema refer to the highest and lowest values of a function on a given interval. To find these, we need to evaluate the function at critical points and endpoints of the interval.
Find the derivative of the function ƒ(x) = x³ - 3x². The derivative, ƒ'(x), will help us identify critical points where the slope is zero or undefined. Calculate ƒ'(x) = 3x² - 6x.
Set the derivative ƒ'(x) = 0 to find critical points. Solve the equation 3x² - 6x = 0. Factor the equation to get 3x(x - 2) = 0, which gives the critical points x = 0 and x = 2.
Evaluate the function ƒ(x) at the critical points and at the endpoints of the interval [-1, 3]. Calculate ƒ(-1), ƒ(0), ƒ(2), and ƒ(3) to find the values of the function at these points.
Compare the values obtained from evaluating the function at the critical points and endpoints. The largest value will be the absolute maximum, and the smallest value will be the absolute minimum on the interval [-1, 3].

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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 are values of x in the domain of a function where the derivative is either zero or undefined. These points are essential for finding local maxima and minima, as they indicate where the function's slope changes. To locate absolute extrema, one must first find these critical points within the given interval.
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Critical Points

Endpoints of the Interval

When determining absolute extrema on a closed interval, it is crucial to evaluate the function at both the critical points and the endpoints of the interval. The absolute maximum or minimum could occur at these endpoints, so they must be included in the analysis to ensure all potential extreme values are considered.
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Estimating the Area Under a Curve with Right Endpoints & Midpoint

First Derivative Test

The First Derivative Test is a method used to determine whether a critical point is a local maximum, local minimum, or neither. By analyzing the sign of the derivative before and after the critical point, one can infer the behavior of the function around that point. This test helps in identifying the nature of the extrema found in the analysis.
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The First Derivative Test: Finding Local Extrema
Related Practice
Textbook Question

Absolute maxima and minima Determine the location and value of the absolute extreme values of ƒ on the given interval, if they exist.


ƒ(x) = x² - 10 on [-2, 3]

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Textbook Question

Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.

lim_x→∞ (log₂ x - log₃ x)

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Textbook Question

17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.


lim_x→ 0⁺ (x - 3 √x) / (x - √x)

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Textbook Question

Increasing and decreasing functions. Find the intervals on which f is increasing and the intervals on which it is decreasing.


f(x) = x² - 2 ln x

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Textbook Question

17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.


lim_x→0 csc 6x sin 7x

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Textbook Question

Second Derivative Test Locate the critical points of the following functions. Then use the Second Derivative Test to determine (if possible) whether they correspond to local maxima or local minima.


f(x) = 6x² - x³

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