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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 4.1.65
Chapter 4, Problem 4.1.65

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.


h(x) = |x + 2| − |x − 3|, −∞ < x < ∞

Verified step by step guidance
1
Step 1: Understand the function h(x) = |x + 2| − |x − 3|. This function involves absolute values, which can be split into piecewise functions based on the critical points where the expressions inside the absolute values change sign.
Step 2: Identify the critical points for the absolute value expressions. The critical points are x = -2 and x = 3, where the expressions inside the absolute values change sign.
Step 3: Break down the function into piecewise components based on the critical points. For x < -2, both expressions are negative, for -2 ≤ x < 3, the first expression is non-negative and the second is negative, and for x ≥ 3, both expressions are non-negative.
Step 4: Graph the piecewise function by evaluating h(x) in each interval: x < -2, -2 ≤ x < 3, and x ≥ 3. This will help visualize the behavior of the function across the entire domain.
Step 5: Determine the extreme values by analyzing the graph and evaluating the function at the critical points and endpoints of the intervals. The extreme values occur where the function reaches its maximum or minimum values within the given domain.

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Key Concepts

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

Absolute Value Functions

Absolute value functions, such as |x + 2| and |x - 3|, measure the distance of a number from zero on the number line, always yielding a non-negative result. These functions create V-shaped graphs and are crucial for understanding how the function h(x) = |x + 2| − |x − 3| behaves, especially at points where the expressions inside the absolute values change sign.
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Graphing Piecewise Functions

Graphing piecewise functions involves plotting different expressions over specified intervals. For h(x) = |x + 2| − |x − 3|, the function can be broken into segments based on the critical points where x + 2 = 0 and x - 3 = 0. Understanding how to graph these segments helps visualize the function's behavior and identify extreme values.
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Finding Extreme Values

Extreme values of a function are its maximum and minimum values within a given interval. To find these for h(x), analyze the critical points where the derivative is zero or undefined, and evaluate the function at these points and endpoints of the interval. This process helps determine where the function reaches its highest or lowest values.
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Related Practice
Textbook Question

22. A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only half as much light per unit area as clear glass does. The total perimeter is fixed. Find the proportions of the window that will admit the most light. Neglect the thickness of the frame.

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Textbook Question

Checking the Mean Value Theorem


Which of the functions in Exercises 7–12 satisfy the hypotheses of the Mean Value Theorem on the given interval, and which do not? Give reasons for your answers.


f(x) = x⁴ᐟ⁵, [0, 1]

Textbook Question

Find values of a and b such that the function


ƒ(𝓍) = (a𝓍 + b) / 𝓍² ―1)


has a local extreme value of 1 at 𝓍 = 3. Is this extreme value a local maximum or a local minimum? Give reasons for your answer.

Textbook Question

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


dy/dx = 3x⁻²ᐟ³, y(−1) = −5

Textbook Question

Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the open intervals on which the functions are differentiable and the graphs are concave up and concave down.

4. y=9/14x^(1/3)(x^2-7)

Textbook Question

Finding Indefinite Integrals


In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.


∫(1 + tan²θ)dθ (Hint:1 + tan²θ = sec²θ)