Skip to main content
Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 4 - Applications of the DerivativeProblem 4.3.35
Chapter 4, Problem 4.3.35

Increasing and decreasing functions. Find the intervals on which f is increasing and the intervals on which it is decreasing.


f(x) = x²/₃ (x²-4)

Verified step by step guidance
1
First, find the derivative of the function f(x) = x^(2/3) * (x^2 - 4). Use the product rule for differentiation, which states that if you have a function h(x) = u(x) * v(x), then h'(x) = u'(x) * v(x) + u(x) * v'(x).
Let u(x) = x^(2/3) and v(x) = x^2 - 4. Find the derivatives u'(x) and v'(x). For u(x) = x^(2/3), use the power rule to get u'(x) = (2/3)x^(-1/3). For v(x) = x^2 - 4, the derivative is v'(x) = 2x.
Apply the product rule: f'(x) = u'(x) * v(x) + u(x) * v'(x). Substitute the derivatives and original functions: f'(x) = (2/3)x^(-1/3) * (x^2 - 4) + x^(2/3) * 2x.
Simplify the expression for f'(x). Combine like terms and simplify the expression to make it easier to analyze the sign of f'(x).
Determine the intervals where f'(x) > 0 (function is increasing) and f'(x) < 0 (function is decreasing). Find the critical points by setting f'(x) = 0 and solving for x. Use these critical points to test intervals on the number line to determine where the function is increasing or decreasing.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
10m
Was this helpful?

Key Concepts

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

Critical Points

Critical points are values of x in a function where the derivative is either zero or undefined. These points are essential for determining where a function changes from increasing to decreasing or vice versa. To find critical points, we first compute the derivative of the function and set it equal to zero, solving for x.
Recommended video:
04:50
Critical Points

First Derivative Test

The First Derivative Test is a method used to determine the behavior of a function at its critical points. By analyzing the sign of the derivative before and after each critical point, we can conclude whether the function is increasing or decreasing in those intervals. If the derivative changes from positive to negative, the function has a local maximum; if it changes from negative to positive, it has a local minimum.
Recommended video:
07:09
The First Derivative Test: Finding Local Extrema

Intervals of Increase and Decrease

Intervals of increase and decrease refer to the ranges of x-values where a function is respectively rising or falling. A function is increasing on an interval if its derivative is positive throughout that interval, and decreasing if the derivative is negative. Identifying these intervals helps in understanding the overall behavior of the function and is crucial for graphing and optimization problems.
Recommended video:
07:32
Determining Where a Function is Increasing & Decreasing
Related Practice
Textbook Question

Metal rain gutters A rain gutter is made from sheets of metal 9 in wide. The gutters have a 3-in base and two 3-in sides, folded up at an angle Θ (see figure). What angle Θ maximizes the cross-sectional area of the gutter? <IMAGE>

288
views
Textbook Question

Explain Rolle’s Theorem with a sketch.

524
views
Textbook Question

{Use of Tech} Critical points and extreme values


a. Find the critical points of the following functions on the given interval. Use a root finder, if necessary.

b. Use a graphing utility to determine whether the critical points correspond to local maxima, local minima, or neither.

c. Find the absolute maximum and minimum values on the given interval, if they exist


h(x) (5-x)/(x² + 2x - 3) on [-10,10]

278
views
Textbook Question

17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.


lim_v→3 (v-1-√(v²-5)) / (v-3)

151
views
Textbook Question

17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.


lim_x→0 (x + cos x)¹/ˣ

190
views
Textbook Question

17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.


lim_x→ 0 (eˣ - x - 1) / 5x²

186
views