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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 4 - Applications of DerivativesProblem 4.1.21
Chapter 4, Problem 4.1.21

Absolute Extrema on Finite Closed Intervals


In Exercises 21–36, find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.


f(x) = (2/3)x − 5, −2 ≤ x ≤ 3

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1
Step 1: Understand the problem. We need to find the absolute maximum and minimum values of the function f(x) = (2/3)x - 5 on the interval [-2, 3]. Absolute extrema occur at critical points or endpoints of the interval.
Step 2: Evaluate the function at the endpoints of the interval. Calculate f(-2) and f(3) to determine the function values at these points. This will help us identify potential extrema.
Step 3: Since f(x) = (2/3)x - 5 is a linear function, it does not have any critical points within the interval. Critical points are found by setting the derivative equal to zero, but the derivative of a linear function is constant and does not equal zero.
Step 4: Compare the function values at the endpoints. The absolute maximum and minimum will be the largest and smallest values obtained from evaluating the function at x = -2 and x = 3.
Step 5: Graph the function f(x) = (2/3)x - 5 over the interval [-2, 3]. Plot the points (-2, f(-2)) and (3, f(3)) on the graph. These points represent the absolute extrema, and their coordinates should be clearly marked.

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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 a function attains on a given interval. To find these values, evaluate the function at critical points and endpoints of the interval. The absolute maximum is the largest value, and the absolute minimum is the smallest value within the interval.
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Finding Extrema Graphically

Critical Points

Critical points are values of x where the derivative of the function is zero or undefined. These points are potential candidates for local extrema. In the context of finding absolute extrema, critical points within the interval, along with endpoints, must be evaluated to determine the function's maximum and minimum values.
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Critical Points

Graphing Linear Functions

Graphing linear functions involves plotting a straight line based on the function's slope and y-intercept. For f(x) = (2/3)x - 5, the slope is 2/3, and the y-intercept is -5. On the interval [-2, 3], plot the line and identify the coordinates of the endpoints to determine where the absolute extrema occur.
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Linearization
Related Practice
Textbook Question

Identifying Extrema


In Exercises 61 and 62, the graph of f' is given. Assume that f is continuous, and determine the x-values corresponding to local minima and local maxima.


Textbook Question

117. Suppose that the second derivative of the function y = f(x) isy" =(x+1)(x-2).

For what x-values does the graph of f have an inflection point?

Textbook Question

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


dv/dt = (1/2)sec t tan t, v(0) = 1

Textbook Question

Finding Critical Points


In Exercises 41–50, determine all critical points and all domain endpoints for each function.


y = x² − 32√x

Textbook Question

Absolute Extrema on Finite Closed Intervals


In Exercises 21–36, find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.


f(t) = 2 − |t|, −1 ≤ t ≤ 3

Textbook Question

Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.

77. y' = cot(θ/2), for 0 < θ < 2π