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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 4.1.63
Chapter 4, Problem 4.1.63

Theory and Examples


[Technology Exercise] Graph the functions in Exercises 63–66. Then find the extreme values of the function on the interval and say where they occur.


f(x) = |x − 2| + |x + 3|, −5 ≤ x ≤ 5

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1
Identify the critical points of the function f(x) = |x - 2| + |x + 3| by considering where the expressions inside the absolute values change sign. These points are x = 2 and x = -3.
Divide the interval [-5, 5] into subintervals based on the critical points: [-5, -3], [-3, 2], and [2, 5].
On each subinterval, rewrite the function f(x) without absolute values. For example, on [-5, -3], f(x) = -(x - 2) - (x + 3).
Calculate the value of f(x) at the endpoints of each subinterval and at the critical points to find potential extreme values.
Compare the values obtained in the previous step to determine the minimum and maximum values of f(x) on the interval [-5, 5], and identify where these extreme values occur.

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Key Concepts

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

Absolute Value Function

The absolute value function, denoted as |x|, measures the distance of a number x from zero on the number line, always yielding a non-negative result. In the function f(x) = |x − 2| + |x + 3|, the absolute value creates a piecewise function that changes its expression based on the sign of the input, which is crucial for graphing and analyzing the function's behavior.
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Piecewise Functions

Piecewise functions are defined by different expressions over different intervals of the domain. For f(x) = |x − 2| + |x + 3|, the function is piecewise because the absolute value expressions change at x = 2 and x = -3, creating different linear segments. Understanding how to break down and analyze these segments is essential for graphing and finding extreme values.
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Extreme Values on an Interval

Extreme values refer to the maximum and minimum values a function attains on a given interval. To find these for f(x) = |x − 2| + |x + 3| on −5 ≤ x ≤ 5, one must evaluate the function at critical points and endpoints. Critical points occur where the derivative is zero or undefined, often at points where the piecewise function changes.
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Related Practice
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.

69. y' = x(x - 3)²

Textbook Question

6. You are planning to close off a corner of the first quadrant with a line segment 20 units long running from (a, 0) to (0,b). Show that the area of the triangle enclosed by the segment is largest when a = b.

Textbook Question

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


dy/dx = 2x − 7, y(2) = 0

Textbook Question

Theory and Examples


Maximum height of a vertically moving body The height of a body moving vertically is given by s = −12gt² + υ₀t + s₀,  g > 0, with s in meters and t in seconds. Find the body’s maximum height.

Textbook Question

Initial Value Problems


Find the curve y = f(x) in the xy-plane that passes through the point (9,4) and whose slope at each point is 3√x.

Textbook Question

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


dy/dx = 1/x² + x, x > 0; y(2) = 1