Ch. 4 - Applications of Derivatives

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

Hass 15th Edition

Ch. 4 - Applications of Derivatives

Problem 4.4.71

Chapter 4, Problem 4.4.71

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.
71. y' = x(x² - 12)

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First, find the second derivative y'' by differentiating y' = x(x² - 12). Use the product rule for differentiation, which states that if you have a function u(x)v(x), its derivative is u'(x)v(x) + u(x)v'(x). Here, let u(x) = x and v(x) = x² - 12.

Differentiate u(x) = x to get u'(x) = 1. Differentiate v(x) = x² - 12 to get v'(x) = 2x.

Apply the product rule: y'' = u'(x)v(x) + u(x)v'(x) = 1(x² - 12) + x(2x). Simplify this expression to find y''.

Now, analyze the critical points and concavity using y'' to understand the general shape of the graph of f(x). Critical points occur where y'' = 0 or is undefined. Determine these points and test intervals around them to see if y'' is positive (indicating concave up) or negative (indicating concave down).

Use the information from the second derivative test to sketch the general shape of the graph of f(x). Consider the behavior at critical points and the concavity to draw a rough sketch of the graph, showing where it is increasing, decreasing, concave up, and concave down.

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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, minima, or points of inflection.

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, suggesting a local minimum; if y'' is negative, the graph is concave down, indicating a local maximum.

Graphing Procedure

The graphing procedure involves several steps to sketch the function based on its derivatives. Steps typically include finding critical points from the first derivative, determining concavity and inflection points from the second derivative, and analyzing the behavior of the function at these points. This systematic approach helps create an accurate representation of the function's overall shape.

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Graphing The Derivative

Related Practice

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.

5. y=x+sin(2x), -2π/3≤x≤2π/3

Textbook Question

Roots (Zeros)

Show that the functions in Exercises 19–26 have exactly one zero in the given interval.

g(t) = 1/(1 − t) + √(1 + t) − 3.1, (−1, 1)

Textbook Question

Root Finding

5. Use Newton's method to find the positive fourth root of 2 by solving the equation x^4 -2 = 0. Start with x_0 = 1 and find x_2.

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.

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.

∫(1 − cot²x) dx

Textbook Question

Initial Value Problems

Solve the initial value problems in Exercises 71–90.

d³y/dx³ = 6; y″(0) = −8, y′(0) = 0, y(0) = 5

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.71. y' = x(x² - 12)

Verified video answer for a similar problem:

Key Concepts

First Derivative

Second Derivative

Graphing Procedure

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.
71. y' = x(x² - 12)