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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 4.1.61
Chapter 4, Problem 4.1.61

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) = x/(x²+9)⁵ on [-2,2]

Verified step by step guidance
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First, understand that absolute maxima and minima 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) = x/(x²+9)⁵. Use the quotient rule, which states that if you have a function g(x)/h(x), its derivative is (g'(x)h(x) - g(x)h'(x))/(h(x))².
Set the derivative equal to zero to find critical points. This involves solving the equation derived from the derivative for x. Critical points occur where the derivative is zero or undefined.
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 = 2.
Compare the values of ƒ(x) at these points to determine the absolute maximum and minimum values on the interval [-2, 2]. The largest value will be the absolute maximum, and the smallest will be the absolute minimum.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Absolute Extrema

Absolute extrema refer to the highest and lowest values of a function on a given interval. To find these values, one must evaluate the function at critical points, where the derivative is zero or undefined, as well as at the endpoints of the interval. The largest of these values is the absolute maximum, while the smallest is the absolute minimum.
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Critical Points

Critical points are values of the independent variable where the derivative of the function is either zero or does not exist. These points are essential for finding absolute extrema, as they indicate where the function may change direction. To locate critical points, one must first compute the derivative of the function and solve for when it equals zero or is undefined.
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Closed Interval Method

The closed interval method is a technique used to find absolute extrema of a function on a closed interval [a, b]. This method involves evaluating the function at the endpoints of the interval and at any critical points found within the interval. The absolute maximum and minimum values are then determined by comparing these function values.
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Related Practice
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) = eˣ(x - 2)²

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Textbook Question

Suppose f is differentiable on (-∞,∞), f(1) = 2, and f'(1) = 3. Find the linear approximation to f at x = 1 and use it to approximate f (1.1).

Textbook Question

Maximum-area rectangles Of all rectangles with a perimeter of 10, which one has the maximum area? (Give the dimensions.)

Textbook Question

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


f(x) = e⁻ˣ²/₂

Textbook Question

{Use of Tech} Finding roots with Newton’s method For the given function f and initial approximation x₀, use Newton’s method to approximate a root of f. Stop calculating approximations when two successive approximations agree to five digits to the right of the decimal point after rounding. Show your work by making a table similar to that in Example 1.


f(x) = x² - 10; x₀ = 3

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Textbook Question

Locating critical points Find the critical points of the following functions. Assume a is a nonzero constant.


ƒ(x) = 3x³ + 3x² / 2 - 2x