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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 54
Chapter 4, Problem 54

First Derivative Test


a. Locate the critical points of f.
b. Use the First Derivative Test to locate the local maximum and minimum values.
c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist).


f(x) = x - 2 tan⁻¹ x on [-√3,√3)

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1
To find the critical points of the function \( f(x) = x - 2 \tan^{-1}(x) \), first compute the derivative \( f'(x) \). The derivative of \( x \) is 1, and the derivative of \( -2 \tan^{-1}(x) \) is \( -\frac{2}{1+x^2} \). Therefore, \( f'(x) = 1 - \frac{2}{1+x^2} \).
Set \( f'(x) = 0 \) to find the critical points. This gives the equation \( 1 - \frac{2}{1+x^2} = 0 \). Solve for \( x \) to find the critical points.
Use the First Derivative Test to determine the nature of each critical point. Evaluate \( f'(x) \) on intervals around each critical point to determine where \( f'(x) \) changes sign. If \( f'(x) \) changes from positive to negative, there is a local maximum. If it changes from negative to positive, there is a local minimum.
To find the absolute maximum and minimum values on the interval \([-\sqrt{3}, \sqrt{3})\), evaluate \( f(x) \) at the critical points and at the endpoints of the interval. Note that the interval is open at \( \sqrt{3} \), so only consider \( x = -\sqrt{3} \) and any critical points within the interval.
Compare the values of \( f(x) \) at these points to determine the absolute maximum and minimum values. The largest value is the absolute maximum, and the smallest value is the absolute minimum.

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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 either zero or undefined. These points are essential for identifying local extrema, 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 values of x that satisfy the condition f'(x) = 0 or where f'(x) does not exist.
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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 conclude that if the derivative changes from positive to negative, the point is a local maximum; if it changes from negative to positive, it is a local minimum. This test provides insight into the function's increasing and decreasing behavior around critical points.
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The First Derivative Test: Finding Local Extrema

Absolute Maximum and Minimum

The absolute maximum and minimum values of a function on a given interval are the highest and lowest values that the function attains within that interval. To find 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 determine the absolute extrema, which are crucial for understanding the overall behavior of the function within the specified range.
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Finding Extrema Graphically Example 4
Related Practice
Textbook Question

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b. At what point of the interval [0, 8] is the instantaneous rate of change equal to the average rate of change?

Textbook Question

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Textbook Question

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a. A bamboo shoot was 500 cm tall at 10:00 A.M. and 515 cm tall at 3:00 P.M. Compute the average growth rate of the bamboo shoot in cm/hr over the period of time from 10:00 A.M. to 3:00 P.M.

Textbook Question

{Use of Tech} Approximating reciprocals To approximate the reciprocal of a number a without using division, we can apply Newton’s method to the function f(x) = 1/x - a. 

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Textbook Question

First Derivative Test


a. Locate the critical points of f.

b. Use the First Derivative Test to locate the local maximum and minimum values.

c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist).


f(x) = x²/(x² - 1) on [-4,4]

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views
Textbook Question

Mean Value Theorem The population of a culture of cells grows according to the function P(t) = 100t / t+1, where t ≥ 0 is measured in weeks.




a. What is the average rate of change in the population over the interval [0, 8]?