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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 4.4.117
Chapter 4, Problem 4.4.117

117. Suppose that the second derivative of the function y = f(x) isy" =(x+1)(x-2).
For what x-values does the graph of f have an inflection point?

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To find the x-values where the graph of f has an inflection point, we need to determine where the second derivative changes sign. An inflection point occurs where the concavity of the function changes.
Start by setting the second derivative equal to zero: y'' = (x+1)(x-2) = 0. Solve this equation to find the critical points.
Factor the equation: (x+1)(x-2) = 0. This gives us two solutions: x = -1 and x = 2.
To confirm these are inflection points, check the sign of the second derivative around these values. Choose test points in the intervals (-\(\text{∞}\), -1), (-1, 2), and (2, \(\text{∞}\)).
Evaluate the sign of y'' at these test points to determine if the sign changes. If the sign changes, then the corresponding x-value is an inflection point.

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Key Concepts

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

Second Derivative Test

The second derivative test is used to determine the concavity of a function and identify inflection points. An inflection point occurs where the second derivative changes sign, indicating a transition from concave up to concave down or vice versa. For the function y = f(x), the inflection points are found by setting the second derivative equal to zero and solving for x.
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The Second Derivative Test: Finding Local Extrema

Factoring Quadratic Expressions

Factoring quadratic expressions involves rewriting them as a product of linear factors. In the given second derivative y'' = (x+1)(x-2), the expression is already factored, indicating potential x-values where the derivative equals zero. These values, x = -1 and x = 2, are critical for determining where the graph might have inflection points.
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Simplifying Trig Expressions

Sign Change Analysis

Sign change analysis involves examining the intervals around the roots of the second derivative to determine where the sign changes. This helps confirm the presence of inflection points. For y'' = (x+1)(x-2), check the sign of y'' in intervals around x = -1 and x = 2 to ensure it changes, confirming these x-values as inflection points.
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Changing Geometries
Related Practice
Textbook Question

Absolute Extrema on Finite Closed Intervals


In Exercises 21–36, find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.


f(x) = (2/3)x − 5, −2 ≤ x ≤ 3

Textbook Question

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


dv/dt = (1/2)sec t tan t, v(0) = 1

Textbook Question

Identifying Extrema


In Exercises 19–40:


a. Find the open intervals on which the function is increasing and those on which it is decreasing.


b. Identify the function’s local extreme values, if any, saying where they occur.


f(x) = x³ / (3x² + 1)

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Textbook Question

Finding Critical Points


In Exercises 41–50, determine all critical points and all domain endpoints for each function.


y = x² − 32√x

Textbook Question

Absolute Extrema on Finite Closed Intervals


In Exercises 21–36, find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.


f(t) = 2 − |t|, −1 ≤ t ≤ 3

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.

80. y' = 1 - cot²θ, for 0 < θ < π