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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 4.3.36
Chapter 4, Problem 4.3.36

Identifying Extrema


In Exercises 19–40:


a. Find the open intervals on which the function is increasing and those on which it is decreasing.


b. Identify the function’s local extreme values, if any, saying where they occur.


f(x) = x³ / (3x² + 1)

Verified step by step guidance
1
To determine where the function \( f(x) = \frac{x^3}{3x^2 + 1} \) is increasing or decreasing, first find the derivative \( f'(x) \). Use the quotient rule: \( f'(x) = \frac{(3x^2)(3x^2 + 1) - (x^3)(6x)}{(3x^2 + 1)^2} \). Simplify the expression to get the derivative.
Set the derivative \( f'(x) \) equal to zero to find critical points. Solve \( (3x^2)(3x^2 + 1) - (x^3)(6x) = 0 \) to find the values of \( x \) where the derivative is zero or undefined.
Determine the sign of \( f'(x) \) on the intervals defined by the critical points. Choose test points in each interval and substitute them into \( f'(x) \) to see if the function is increasing (\( f'(x) > 0 \)) or decreasing (\( f'(x) < 0 \)).
Identify the local extrema by analyzing the sign changes of \( f'(x) \). If \( f'(x) \) changes from positive to negative at a critical point, there is a local maximum. If it changes from negative to positive, there is a local minimum.
Finally, state the open intervals where the function is increasing or decreasing and identify any local extrema, specifying their locations based on the critical points and the behavior of \( f'(x) \).

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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 provides information about the function's rate of change. To find where a function is increasing or decreasing, calculate its derivative and identify critical points where the derivative is zero or undefined. These points are potential locations for local extrema and help determine intervals of increase or decrease.
Recommended video:
04:50
Critical Points

First Derivative Test

The First Derivative Test is used to determine whether a critical point is a local maximum, minimum, or neither. By analyzing the sign of the derivative before and after a critical point, one can ascertain the behavior of the function: 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.
Recommended video:
07:09
The First Derivative Test: Finding Local Extrema

Interval Testing

Interval testing involves selecting test points from intervals determined by critical points to evaluate the sign of the derivative. This process helps identify where the function is increasing or decreasing. By substituting these test points into the derivative, one can confirm the function's behavior over each interval, aiding in the identification of local extrema.
Recommended video:
07:09
The First Derivative Test: Finding Local Extrema
Related Practice
Textbook Question

117. Suppose that the second derivative of the function y = f(x) isy" =(x+1)(x-2).

For what x-values does the graph of f have an inflection point?

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Textbook Question

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


dv/dt = (1/2)sec t tan t, v(0) = 1

13
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Textbook Question

Finding Critical Points


In Exercises 41–50, determine all critical points and all domain endpoints for each function.


y = x² + 2/x

159
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Textbook Question

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

206
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Textbook Question

Solve the initial value problems in Exercises 71–90.

y⁽⁴⁾ = −sin t + cos t;

y′′′(0) =7, y′′(0) = y′(0) = −1, y(0) = 0

18
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Textbook Question

Checking the Mean Value Theorem


Find the value or values of c that satisfy the equation (f(b) − f(a)) / (b − a) = f′(c) in the conclusion of the Mean Value Theorem for the functions and intervals in Exercises 1–6.


g(x) = {x³, −2 ≤ x ≤ 0

x², 0 < x ≤ 2

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