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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 4.4.69
Chapter 4, Problem 4.4.69

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.
69. y' = x(x - 3)²

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Step 1: Start by finding the second derivative y''. To do this, differentiate the given first derivative y' = x(x - 3)² using the product rule. Recall that the product rule states: (uv)' = u'v + uv'. Here, u = x and v = (x - 3)².
Step 2: Differentiate u = x to get u' = 1. Then, differentiate v = (x - 3)² using the chain rule. The chain rule states: (g(h(x)))' = g'(h(x)) * h'(x). Here, g(h) = h² and h(x) = x - 3. So, v' = 2(x - 3)(1) = 2(x - 3).
Step 3: Apply the product rule to compute y''. Substitute u, u', v, and v' into the formula: y'' = u'v + uv'. This becomes y'' = (1)(x - 3)² + (x)(2(x - 3)). Simplify this expression to get the second derivative.
Step 4: Analyze the critical points and concavity using y'' to determine the general shape of the graph. Set y'' = 0 to find inflection points, and test intervals around these points to determine where the graph is concave up (y'' > 0) or concave down (y'' < 0).
Step 5: Combine the information from y' and y'' to sketch the general shape of the graph. Use the critical points from y' (where y' = 0) to identify maxima, minima, or points of inflection, and use the concavity information from y'' to refine the sketch of the graph.

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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 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. If y'' is positive, the graph is concave up, and if y'' is negative, the graph is concave down. This information is crucial for sketching the general shape of the function.
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The Second Derivative Test: Finding Local Extrema

Graphing Procedure

The graphing procedure involves several steps to sketch the graph of a 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 endpoints or asymptotes. 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

In Exercises 1–10, find the extreme values (absolute and local) of the function over its natural domain, and where they occur.

__________

y = √ 3 + 2𝓍 ―𝓍²

Textbook Question

6. You are planning to close off a corner of the first quadrant with a line segment 20 units long running from (a, 0) to (0,b). Show that the area of the triangle enclosed by the segment is largest when a = b.

Textbook Question

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


dy/dx = 2x − 7, y(2) = 0

Textbook Question

Theory and Examples


[Technology Exercise] Graph the functions in Exercises 63–66. Then find the extreme values of the function on the interval and say where they occur.


f(x) = |x − 2| + |x + 3|, −5 ≤ x ≤ 5

Textbook Question

Initial Value Problems


Find the curve y = f(x) in the xy-plane that passes through the point (9,4) and whose slope at each point is 3√x.

Textbook Question

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


dy/dx = 1/x² + x, x > 0; y(2) = 1