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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 4.4.109
Chapter 4, Problem 4.4.109

109. Suppose the derivative of the function y = f(x) is
y'=(x-1)^2(x-2).
At what points, if any, does the graph of f have a local minimum, local maximum, or
point of inflection? (Hint: Draw the sign pattern for y'.)

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1
Step 1: Identify critical points by setting the derivative y' = (x-1)^2(x-2) equal to zero. Solve for x to find the values where the slope of the tangent line is zero or undefined. This gives x = 1 and x = 2 as potential critical points.
Step 2: Analyze the sign pattern of y' by testing intervals around the critical points. Divide the x-axis into intervals based on the critical points: (-∞, 1), (1, 2), and (2, ∞). Choose test points within each interval and substitute them into y' to determine whether y' is positive or negative in each interval.
Step 3: Determine the behavior of the graph at x = 1 and x = 2. For x = 1, note that (x-1)^2 is always non-negative, so y' does not change sign at x = 1. This indicates that x = 1 is not a local maximum or minimum but could be a point of inflection. For x = 2, check if y' changes sign from positive to negative or vice versa to identify a local maximum or minimum.
Step 4: Investigate the second derivative y'' to confirm points of inflection. Compute y'' by differentiating y' = (x-1)^2(x-2) and analyze its sign changes. Points where y'' changes sign indicate points of inflection.
Step 5: Summarize the findings. Based on the sign pattern of y' and the behavior of y'' at x = 1 and x = 2, classify each critical point as a local minimum, local maximum, or point of inflection. Ensure the reasoning aligns with the sign changes observed in y' and y''.

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Key Concepts

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

Derivative and Critical Points

The derivative of a function, denoted as y' or f'(x), represents the rate of change of the function at any point. Critical points occur where the derivative is zero or undefined, indicating potential locations for local maxima, minima, or points of inflection. Analyzing these points helps determine the behavior of the function around them.
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Sign Chart Analysis

A sign chart is a visual tool used to determine the intervals where a function's derivative is positive, negative, or zero. By evaluating the sign of the derivative around critical points, one can infer whether the function is increasing or decreasing, which aids in identifying local maxima and minima, as well as points of inflection.
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Derivatives Applied To Acceleration Example 2

Local Extrema and Points of Inflection

Local extrema refer to points where a function reaches a local maximum or minimum, determined by the sign of the derivative. A point of inflection occurs where the concavity of the function changes, typically identified by a change in the sign of the second derivative. Understanding these concepts is crucial for analyzing the overall shape and behavior of the graph of the function.
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Related Practice
Textbook Question

Root Finding

2. Use Newton's method to estimate the one real solution of x^3 +3x + 1 = 0. Start with x_0 = 0 and then find x_2.

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.

82. y' = sin t, for 0 ≤ t ≤ 2π

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.


∫(2x³ − 5x + 7) dx

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) = x²ᐟ³, [−1, 8]

Textbook Question

Roots (Zeros)


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


f(x) = x⁴ + 3x + 1, [−2, −1]

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