Ch. 4 - Applications of Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 4 - Applications of Derivatives

Problem 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?

Verified step by step guidance

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.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

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.

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.

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.

Recommended video:

06:35

Changing Geometries

Related Practice

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.

∫(sin2x − csc²x)dx

views

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

views

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)

159

views

Textbook Question

"Roots (Zeros) Show that the functions in Exercises 19–26 have exactly one zero

206

views

Textbook Question

Checking the Mean Value Theorem

Find the value or values of c that satisfy the equation (f(b) − f(a)) / (b − a) = f′(c) in the conclusion of the Mean Value Theorem for the functions and intervals in Exercises 1–6.

g(x) = {x³, −2 ≤ x ≤ 0

x², 0 < x ≤ 2

245

views

Textbook Question

Finding Critical Points

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

y = x² − 32√x

200

views