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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 4.4.85
Chapter 4, Problem 4.4.85

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.
85. y' = x^(-2/3) (x - 1)

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Step 1: Compute the second derivative y'' by applying the product rule to y' = x^(-2/3) * (x - 1). Recall the product rule: (uv)' = u'v + uv'. Here, u = x^(-2/3) and v = (x - 1). Differentiate each term separately.
Step 2: For u = x^(-2/3), compute u' using the power rule: d/dx[x^n] = n*x^(n-1). This gives u' = (-2/3)*x^(-5/3). For v = (x - 1), compute v' as the derivative of (x - 1), which is simply 1.
Step 3: Substitute u, u', v, and v' into the product rule formula: y'' = u'v + uv'. This results in y'' = [(-2/3)*x^(-5/3)*(x - 1)] + [x^(-2/3)*1]. Simplify the expression to combine like terms.
Step 4: Analyze the critical points and inflection points by setting y'' = 0 and solving for x. These points will help determine the concavity and general shape of the graph of f(x). Additionally, consider the behavior of y'' as x approaches 0 and infinity to understand the graph's asymptotic behavior.
Step 5: Use the information from y'' and y' to sketch the general shape of the graph of f(x). Identify intervals where the graph is concave up or concave down, and mark any inflection points. Combine this with the behavior of y' to determine increasing or decreasing intervals.

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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 its variable. It provides information about the slope of the tangent line to the graph of the function at any given point. Understanding the first derivative is crucial for determining 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. A positive second derivative suggests the graph is concave up, while a negative second derivative indicates concave down, which is essential for sketching the function's general shape.
Recommended video:
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The Second Derivative Test: Finding Local Extrema

Graphing Procedure

The graphing procedure involves analyzing the first and second derivatives to sketch the function's graph. Steps typically include identifying critical points, determining intervals of increase or decrease, and assessing concavity. This systematic approach helps in visualizing the behavior of the function, including where it may have peaks, valleys, or inflection points.
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Graphing The Derivative
Related Practice
Textbook Question

Finding Extrema from Graphs


In Exercises 11–14, match the table with a graph.


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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/x² − x² − 1/3) dx

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Textbook Question

41. Among all triangles in the first quadrant formed by the x-axis, the y-axis, and tangent lines to the graph of y=3x-x^2, what is the smallest possible area?

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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) = x⁴ᐟ⁵, [0, 1]

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Textbook Question

Find values of a and b such that the function


ƒ(𝓍) = (a𝓍 + b) / 𝓍² ―1)


has a local extreme value of 1 at 𝓍 = 3. Is this extreme value a local maximum or a local minimum? Give reasons for your answer.

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Textbook Question

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


dy/dx = 3x⁻²ᐟ³, y(−1) = −5

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