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.
∫(2x³ − 5x + 7) dx
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 4.4.74
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
Verified step by step guidance1
Step 1: Start by finding the second derivative y''. To do this, apply the product rule to y' = (x² - 2x)(x - 5)². Recall the product rule: if y' = u(x)v(x), then y'' = u'(x)v(x) + u(x)v'(x). Here, u(x) = (x² - 2x) and v(x) = (x - 5)².
Step 2: Compute the derivative of u(x) = (x² - 2x). Use the power rule to find u'(x) = 2x - 2.
Step 3: Compute the derivative of v(x) = (x - 5)². Use the chain rule to find v'(x) = 2(x - 5).
Step 4: Substitute u(x), u'(x), v(x), and v'(x) into the product rule formula for y''. This gives y'' = [(2x - 2)(x - 5)²] + [(x² - 2x)(2(x - 5))]. Simplify this expression by expanding and combining like terms.
Step 5: Use the second derivative y'' to analyze the concavity of the graph of f(x). Identify where y'' > 0 (indicating concave up) and y'' < 0 (indicating concave down). Combine this information with the critical points from y' = 0 to sketch the general shape of the graph of f(x).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
First Derivative
The first derivative of a function, denoted as y' or f'(x), represents the rate of change of the function with respect to x. It provides information about the slope of the tangent line to the graph of the function at any given point. Analyzing the first derivative helps identify critical points, where the function may have local maxima or minima.
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The First Derivative Test: Finding Local Extrema
Second Derivative
The second derivative, denoted as y'' or f''(x), is the derivative of the first derivative. It indicates the rate of change of the slope of the function, providing insights into the concavity of the graph. If y'' is positive, the graph is concave up, and if y'' is negative, the graph is concave down. This information is crucial for sketching the general shape of the function.
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06:02The Second Derivative Test: Finding Local Extrema
Graphing Procedure Steps
The graphing procedure involves several steps to analyze the behavior of a function based on its derivatives. Steps typically include finding critical points, determining intervals of increase and decrease, analyzing concavity using the second derivative, and identifying points of inflection. These steps collectively help in sketching an accurate representation of the function's graph.
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06:15Graphing The Derivative
Related Practice
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) = x²ᐟ³, [−1, 8]
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
Textbook Question
119. Find the values of constants a, b, and c such that the graph of y = ax^3 + bx^2 + cx has a
local maximum at x = 3, local minimum at x =- 1, and inflection point at (1, 11).
Textbook Question
Sketch the graphs of the rational functions in Exercises 53–60.
𝓍⁴ ― 1
y = ------------------
𝓍²
Textbook Question
"Roots (Zeros) Show that the functions in Exercises 19–26 have exactly one zero