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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 4.1.57
Chapter 4, Problem 4.1.57

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) = 2x³ - 15x² + 24x on [0,5]

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First, find the derivative of the function ƒ(x) = 2x³ - 15x² + 24x to determine the critical points. The derivative, ƒ'(x), is calculated using the power rule: ƒ'(x) = 6x² - 30x + 24.
Set the derivative ƒ'(x) equal to zero to find the critical points: 6x² - 30x + 24 = 0. Solve this quadratic equation using the quadratic formula or factoring to find the values of x.
Evaluate the function ƒ(x) at the critical points found in the previous step, as well as at the endpoints of the interval [0,5]. This means calculating ƒ(0), ƒ(5), and ƒ at each critical point within the interval.
Compare the values of ƒ(x) 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 [0,5].
Conclude by stating the location (x-value) and the value of the absolute maximum and minimum based on the comparisons made in the previous step.

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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 on a closed interval, one must evaluate the function at these critical points as well as at the endpoints of the interval.
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Critical Points

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. If the derivative changes from positive to negative, the critical point is a local maximum; if it changes from negative to positive, it is a local minimum.
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The First Derivative Test: Finding Local Extrema

Evaluating Endpoints

When finding absolute extrema on a closed interval, it is crucial to evaluate the function at the endpoints of the interval in addition to the critical points. The absolute maximum or minimum could occur at these endpoints, especially if the function is not continuous or differentiable throughout the interval. Thus, comparing the function values at critical points and endpoints ensures that all potential extreme values are considered.
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Related Practice
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Textbook Question

Use the following graphs to identify the points (if any) on the interval [a, b] at which the function has an absolute maximum or an absolute minimum value <IMAGE>

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Textbook Question

Graphing functions Use the guidelines of this section to make a complete graph of f.

f(x) = x³ - 6x² + 9x