Textbook Question
Initial Value Problems
Solve the initial value problems in Exercises 71–90.
d²r/dt² = 2/t³; dr/dt|ₜ ₌ ₁ =1, r(1) = 1
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Ch. 4 - Applications of Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
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
Verified step by step guidance1
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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Average Value of a Function
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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05:36Piecewise Functions
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
Finding Extrema from Graphs
In Exercises 1–6, determine from the graph whether the function has any absolute extreme values on [a, b]. Then explain how your answer is consistent with Theorem 1.
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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.
74. y' = (x² - 2x)(x - 5)²
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Textbook Question
Finding Extreme Values
In Exercises 1–10, find the extreme values (absolute and local) of the function over its natural domain, and where they occur.
y = 𝓍³ + 𝓍² ― 8𝓍 + 5
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Textbook Question
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 < ∞
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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)
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