Textbook Question
104–107. Functions from derivatives Find the function f with the following properties.
h'(x) = (x⁴ -2) /(1 + x²) ; h (1) = -(2/3)
Ch. 4 - Applications of the Derivative
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 4, Problem 4.R.5b
Use the graphs of ƒ' and ƒ" to complete the following steps. <IMAGE>
b. Determine the locations of the inflection points of f and the intervals on which f is concave up or concave down.
Verified step by step guidance1
First, understand that inflection points occur where the second derivative, ƒ'', changes sign. This means you need to analyze the graph of ƒ'' to identify where it crosses the x-axis.
Look at the graph of ƒ'' and identify the x-values where the graph crosses the x-axis. These x-values are potential inflection points because they indicate a change in concavity.
To determine the intervals of concavity, observe the sign of ƒ'' on either side of each inflection point. If ƒ'' is positive, the function ƒ is concave up; if ƒ'' is negative, the function ƒ is concave down.
List the intervals where ƒ'' is positive and label them as concave up intervals. Similarly, list the intervals where ƒ'' is negative and label them as concave down intervals.
Verify the inflection points by ensuring that the sign of ƒ'' changes at these points. This confirms that the function ƒ changes concavity at these x-values.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Inflection Points
Inflection points are points on the graph of a function where the concavity changes. This occurs when the second derivative of the function, f'', is equal to zero or undefined. Identifying these points is crucial for understanding the behavior of the function, as they indicate where the graph shifts from being concave up to concave down, or vice versa.
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Critical Points
Concavity
Concavity refers to the direction in which a curve bends. A function is concave up on an interval if its second derivative, f'', is positive, indicating that the slope of the tangent line is increasing. Conversely, a function is concave down if f'' is negative, meaning the slope of the tangent line is decreasing. Understanding concavity helps in analyzing the overall shape of the graph.
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05:59Determining Concavity Given a Function
Second Derivative Test
The second derivative test is a method used to determine the concavity of a function and locate inflection points. By examining the sign of the second derivative, f'', one can ascertain whether the function is concave up or down. If f'' changes sign at a point, that point is an inflection point, providing valuable information about the function's behavior in that vicinity.
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06:02The Second Derivative Test: Finding Local Extrema
Related Practice
Textbook Question
Find the critical points of the following functions on the given intervals. Identify the absolute maximum and absolute minimum values (if they exist).
ƒ(x) = 4x¹⸍² - x⁵⸍² on [0, 4]
Textbook Question
90–103. Indefinite integrals Determine the following indefinite integrals.
∫(1 + 3 cosΘ) dΘ
Textbook Question
82–89. Comparing growth rates Determine which of the two functions grows faster, or state that they have comparable growth rates.
2ˣ and 4ˣ⸍²
Textbook Question
24–34. Curve sketching Use the guidelines given in Section 4.4 to make a complete graph of the following functions on their domains or on the given interval. Use a graphing utility to check your work.
{Use of Tech} ƒ(x) = x (x -1)e⁻ˣ
Textbook Question
Optimization A right triangle has legs of length h and r and a hypotenuse of length 4 (see figure). It is revolved about the leg of length h to sweep out a right circular cone. What values of h and r maximize the volume of the cone? (Volume of a cone = (1/3) πr²h.) <IMAGE>
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