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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 4.1.79
Chapter 4, Problem 4.1.79

{Use of Tech} Absolute maxima and minima


a. Find the critical points of f on the given interval.
b. Determine the absolute extreme values of f on the given interval.
c. Use a graphing utility to confirm your conclusions.


f(x) = 2ᶻ sin x on [-2,6]

Verified step by step guidance
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To find the critical points of the function f(x) = 2^x sin(x) on the interval [-2, 6], first compute the derivative f'(x). Use the product rule: if u(x) = 2^x and v(x) = sin(x), then f'(x) = u'(x)v(x) + u(x)v'(x).
Calculate the derivatives: u'(x) = 2^x ln(2) and v'(x) = cos(x). Substitute these into the product rule to get f'(x) = 2^x ln(2) sin(x) + 2^x cos(x).
Set the derivative f'(x) to zero to find the critical points: 2^x ln(2) sin(x) + 2^x cos(x) = 0. Factor out 2^x to simplify: 2^x (ln(2) sin(x) + cos(x)) = 0. Since 2^x is never zero, solve ln(2) sin(x) + cos(x) = 0 for x in the interval [-2, 6].
Evaluate f(x) at the critical points found in the previous step, as well as at the endpoints of the interval, x = -2 and x = 6, to determine the absolute maximum and minimum values.
Use a graphing utility to plot f(x) = 2^x sin(x) over the interval [-2, 6] and visually confirm the locations and values of the absolute maxima and minima identified in the previous steps.

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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 of a function occur where its derivative is 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 the condition.
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Absolute Extrema

Absolute extrema refer to the highest and lowest values of a function on a given interval. To determine these values, one must evaluate the function at its critical points and at the endpoints of the interval. The largest value found will be the absolute maximum, while the smallest will be the absolute minimum, providing a complete picture of the function's behavior over the specified range.
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Finding Extrema Graphically

Graphing Utility

A graphing utility is a software tool or calculator that allows users to visualize mathematical functions and their properties. By plotting the function, one can easily observe its critical points, local maxima, and minima, as well as confirm analytical findings. This visual representation aids in understanding the function's overall behavior and verifying calculations made during the analysis.
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Related Practice
Textbook Question

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ƒ(x) = 3x² - 4x + 2

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lim_x→0 (x + cos x)¹/ˣ

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p(x) = 3 sec² x

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Particular antiderivatives For the following functions f, find the antiderivative F that satisfies the given condition.


f(x) = 8x³ + sin x; F(0) = 2

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{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) = cos⁻¹ x - x; x₀ = 0.75

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Textbook Question

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lim_x→ ∞ (e³ˣ ) / (3e³ˣ + 5)

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