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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 4.1.47
Chapter 4, Problem 4.1.47

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) = 3x⁵ - 25x³ + 60x on [-2,3]

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First, find the critical points of the function ƒ(x) = 3x⁵ - 25x³ + 60x by taking its derivative and setting it equal to zero. The derivative is ƒ'(x) = 15x⁴ - 75x² + 60.
Solve the equation 15x⁴ - 75x² + 60 = 0 to find the critical points. This can be done by factoring or using the quadratic formula on the resulting polynomial.
Evaluate the function ƒ(x) at the critical points found in the previous step, as well as at the endpoints of the interval, x = -2 and x = 3.
Compare the values of ƒ(x) at the critical points and the endpoints to determine which is the absolute maximum and which is the absolute minimum on the interval [-2, 3].
Conclude by stating the location (x-values) and the corresponding values of the absolute maximum and minimum of the function on the given interval.

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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 has significant changes in behavior at the edges of the interval. Thus, comparing the function values at critical points and endpoints ensures all potential extrema are considered.
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