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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 4 - Applications of the DerivativeProblema 4.R.6
Capitolo 4, Problema 4.R.6

Find the critical points of the following functions on the given intervals. Identify the absolute maximum and absolute minimum values (if they exist).
ƒ(x) = x³ - 6x² on [-1, 5]

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First, find the derivative of the function ƒ(x) = x³ - 6x². The derivative, ƒ'(x), will help us identify the critical points. Use the power rule to differentiate: ƒ'(x) = 3x² - 12x.
Set the derivative ƒ'(x) = 3x² - 12x equal to zero to find the critical points. Solve the equation 3x² - 12x = 0 for x. Factor the equation: 3x(x - 4) = 0, which gives the solutions x = 0 and x = 4.
Evaluate the function ƒ(x) at the critical points and at the endpoints of the interval [-1, 5]. Calculate ƒ(-1), ƒ(0), ƒ(4), and ƒ(5) to determine the function values at these points.
Compare the values of ƒ(x) at the critical points and endpoints to identify the absolute maximum and minimum values on the interval [-1, 5]. The largest value will be the absolute maximum, and the smallest value will be the absolute minimum.
Conclude by stating the absolute maximum and minimum values, if they exist, based on the evaluations from the previous step. Ensure to check that these values fall within the given interval.

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Critical Points

Critical points of a function occur where its derivative is either zero or undefined. These points are essential for identifying local maxima and minima, as they represent potential locations where the function's behavior changes. To find critical points, one must first compute the derivative of the function and solve for the values of x that satisfy these conditions.
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Critical Points

Absolute Maximum and Minimum

The absolute maximum and minimum values of a function on a closed interval are the highest and lowest values that the function attains within that interval. To determine these values, one must evaluate the function at its critical points and at the endpoints of the interval. The largest and smallest of these values will indicate the absolute maximum and minimum, respectively.
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Finding Extrema Graphically Example 4

Closed Interval

A closed interval, denoted as [a, b], includes all numbers between a and b, as well as the endpoints a and b themselves. This is significant in calculus because the Extreme Value Theorem states that a continuous function on a closed interval must attain both an absolute maximum and minimum. Thus, when analyzing functions over closed intervals, one must consider the behavior of the function at both the endpoints and any critical points within the interval.
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Finding Global Extrema (Extreme Value Theorem)
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