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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 4.3
Chapter 4, Problem 4.3

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


y = 𝓍³ + 𝓍² ― 8𝓍 + 5

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1
First, find the derivative of the function \( y = x^3 + x^2 - 8x + 5 \) to determine the critical points. The derivative is \( y' = 3x^2 + 2x - 8 \).
Set the derivative equal to zero to find the critical points: \( 3x^2 + 2x - 8 = 0 \). Solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = 2 \), and \( c = -8 \).
Calculate the discriminant \( b^2 - 4ac \) to determine the nature of the roots. If the discriminant is positive, there are two distinct real roots, which are the critical points.
Evaluate the second derivative \( y'' = 6x + 2 \) at each critical point to determine the concavity and identify whether each critical point is a local maximum, local minimum, or a point of inflection.
Finally, analyze the behavior of the function as \( x \to \pm \infty \) to determine if there are any absolute extreme values over the natural domain. Consider the end behavior of the cubic function to conclude about absolute extrema.

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Key Concepts

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

Critical Points

Critical points of a function occur where its derivative is zero or undefined. These points are potential locations for local extrema. To find them, compute the derivative of the function and solve for values of x where the derivative equals zero or does not exist. In this problem, finding the derivative of y = xΒ³ + xΒ² - 8x + 5 is essential to identify critical points.
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Critical Points

First Derivative Test

The First Derivative Test helps determine whether a critical point is a local maximum, minimum, or neither. By analyzing the sign changes of the derivative around the critical points, one can infer the nature of the extrema. If the derivative changes from positive to negative, the point is a local maximum; if it changes from negative to positive, it's a local minimum.
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The First Derivative Test: Finding Local Extrema

Second Derivative Test

The Second Derivative Test provides another method to classify critical points. By evaluating the second derivative at a critical point, one can determine concavity: if the second derivative is positive, the function is concave up, indicating a local minimum; if negative, concave down, indicating a local maximum. This test complements the First Derivative Test for confirming extrema.
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The Second Derivative Test: Finding Local Extrema
Related Practice
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, [βˆ’1, 8]

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.

74. y' = (xΒ² - 2x)(x - 5)Β²

Textbook Question

119. Find the values of constants a, b, and c such that the graph of y = ax^3 + bx^2 + cx has a

local maximum at x = 3, local minimum at x =- 1, and inflection point at (1, 11).

Textbook Question

Sketch the graphs of the rational functions in Exercises 53–60.


𝓍⁴ ― 1

y = ------------------

𝓍²

Textbook Question

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

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.


∫(8y βˆ’ 2 / y¹ᐟ⁴) dy