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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 4.1.59
Capitolo 4, Problema 4.1.59

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) = (4x³/3) + 5x² - 6x on [0,5]

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1
First, find the derivative of the function ƒ(x) = \( \frac{4x^3}{3} + 5x^2 - 6x \). This will help us identify critical points where the slope of the tangent is zero or undefined.
Set the derivative equal to zero to find the critical points. Solve the equation \( \frac{d}{dx} \left( \frac{4x^3}{3} + 5x^2 - 6x \right) = 0 \).
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 will help determine the potential locations of absolute extrema.
Compare the values of ƒ(x) at the critical points and the endpoints. The largest value will be the absolute maximum, and the smallest value will be the absolute minimum on the interval.
Conclude by stating the location and value of the absolute maximum and minimum based on the evaluations from the previous step.

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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

Endpoints of the Interval

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 no critical points within the interval. Thus, both critical points and endpoints must be considered to ensure all potential extrema are identified.
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Estimating the Area Under a Curve with Right Endpoints & Midpoint
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