BuyFindarrow_forward

Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

Solutions

Chapter
Section
BuyFindarrow_forward

Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

In Problems 31-36, both a function and its derivative are given. Use them to find critical values, critical points, intervals on which the function is increasing and decreasing, relative maxima, relative minima, and horizontal points of inflection; sketch the graph of each function.

y = x 3 ( x 5 ) 2 27    d y d x = 5 x 2 ( x 3 ) ( x 5 ) 27

To determine

To calculate: The critical values, critical points, the relative maxima, relative minima, or horizontal points of inflection of the function y=x3(x5)227 and the derivative dydx=5x2(x3)(x5)27 and also sketch the graph.

Explanation

Given Information:

The provided function is y=x3(x5)227 and the derivative is dydx=5x2(x3)(x5)27.

Formula Used:

The simple power rule of derivative,

ddx(xn)=nxn1

To obtain the relative maxima and minima of a function,

1. Find the first derivative of the function.

2. Set the derivative equal to 0 to obtain the critical points.

3. Evaluate f(x) at values of x to the left and one to the right of each critical point to develop a sign diagram.

(a) If f(x)>0 to the left and f(x)<0 to the right of the critical value, the critical point is a relative maximum point.

(b) If f(x)<0 to the left and f(x)>0 to the right of the critical value, the critical point is a relative minimum point.

Calculation:

Consider the provided function y=x3(x5)227, and the derivative dydx=5x2(x3)(x5)27,

Now, equate the derivative to 0.

dydx=05x2(x3)(x5)27=0

Evaluate the values of x from the equation:

5x2(x3)(x5)27=0x=0,5,3

Hence, the value(s) of x are x=0,5,3.

Evaluate the values of the original functions with the critical values:

Substitute 0 for x in the function y=x3(x5)227.

y=(0)3(05)227=027=0

Hence, (0,0) is a critical point.

Put x=3 in the equation y=x3(x5)227,

y=(3)3(35)227=27(4)27=4

Hence, (3,4) is a critical point.

Put x=5 in the equation y=x3(x5)227,

y=(5)3(55)227=125(0)27=0

Hence, (5,0) is a critical point

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Chapter 10 Solutions

Show all chapter solutions add
Sect-10.1 P-7ESect-10.1 P-8ESect-10.1 P-9ESect-10.1 P-10ESect-10.1 P-11ESect-10.1 P-12ESect-10.1 P-13ESect-10.1 P-14ESect-10.1 P-15ESect-10.1 P-16ESect-10.1 P-17ESect-10.1 P-18ESect-10.1 P-19ESect-10.1 P-20ESect-10.1 P-21ESect-10.1 P-22ESect-10.1 P-23ESect-10.1 P-24ESect-10.1 P-25ESect-10.1 P-26ESect-10.1 P-27ESect-10.1 P-28ESect-10.1 P-29ESect-10.1 P-30ESect-10.1 P-31ESect-10.1 P-32ESect-10.1 P-33ESect-10.1 P-34ESect-10.1 P-35ESect-10.1 P-36ESect-10.1 P-37ESect-10.1 P-38ESect-10.1 P-39ESect-10.1 P-40ESect-10.1 P-41ESect-10.1 P-42ESect-10.1 P-43ESect-10.1 P-44ESect-10.1 P-45ESect-10.1 P-46ESect-10.1 P-47ESect-10.1 P-48ESect-10.1 P-49ESect-10.1 P-50ESect-10.1 P-51ESect-10.1 P-52ESect-10.1 P-53ESect-10.1 P-54ESect-10.1 P-55ESect-10.1 P-56ESect-10.1 P-57ESect-10.1 P-58ESect-10.1 P-59ESect-10.1 P-60ESect-10.1 P-61ESect-10.1 P-62ESect-10.1 P-63ESect-10.1 P-64ESect-10.1 P-65ESect-10.2 P-1CPSect-10.2 P-2CPSect-10.2 P-3CPSect-10.2 P-4CPSect-10.2 P-1ESect-10.2 P-2ESect-10.2 P-3ESect-10.2 P-4ESect-10.2 P-5ESect-10.2 P-6ESect-10.2 P-7ESect-10.2 P-8ESect-10.2 P-9ESect-10.2 P-10ESect-10.2 P-11ESect-10.2 P-12ESect-10.2 P-13ESect-10.2 P-14ESect-10.2 P-15ESect-10.2 P-16ESect-10.2 P-17ESect-10.2 P-18ESect-10.2 P-19ESect-10.2 P-20ESect-10.2 P-21ESect-10.2 P-22ESect-10.2 P-23ESect-10.2 P-24ESect-10.2 P-25ESect-10.2 P-26ESect-10.2 P-27ESect-10.2 P-28ESect-10.2 P-29ESect-10.2 P-30ESect-10.2 P-31ESect-10.2 P-32ESect-10.2 P-33ESect-10.2 P-34ESect-10.2 P-35ESect-10.2 P-36ESect-10.2 P-37ESect-10.2 P-38ESect-10.2 P-39ESect-10.3 P-1CPSect-10.3 P-2CPSect-10.3 P-3CPSect-10.3 P-4CPSect-10.3 P-1ESect-10.3 P-2ESect-10.3 P-3ESect-10.3 P-5ESect-10.3 P-6ESect-10.3 P-7ESect-10.3 P-8ESect-10.3 P-9ESect-10.3 P-10ESect-10.3 P-11ESect-10.3 P-12ESect-10.3 P-13ESect-10.3 P-14ESect-10.3 P-15ESect-10.3 P-16ESect-10.3 P-17ESect-10.3 P-18ESect-10.3 P-19ESect-10.3 P-20ESect-10.3 P-21ESect-10.3 P-22ESect-10.3 P-23ESect-10.3 P-24ESect-10.3 P-25ESect-10.3 P-26ESect-10.3 P-27ESect-10.3 P-28ESect-10.3 P-29ESect-10.3 P-30ESect-10.3 P-33ESect-10.3 P-34ESect-10.3 P-35ESect-10.3 P-36ESect-10.3 P-37ESect-10.3 P-38ESect-10.3 P-39ESect-10.3 P-40ESect-10.3 P-41ESect-10.3 P-42ESect-10.3 P-43ESect-10.3 P-47ESect-10.3 P-48ESect-10.3 P-49ESect-10.4 P-1CPSect-10.4 P-2CPSect-10.4 P-3CPSect-10.4 P-1ESect-10.4 P-2ESect-10.4 P-3ESect-10.4 P-4ESect-10.4 P-5ESect-10.4 P-6ESect-10.4 P-7ESect-10.4 P-8ESect-10.4 P-9ESect-10.4 P-10ESect-10.4 P-11ESect-10.4 P-12ESect-10.4 P-13ESect-10.4 P-14ESect-10.4 P-15ESect-10.4 P-16ESect-10.4 P-17ESect-10.4 P-18ESect-10.4 P-19ESect-10.4 P-20ESect-10.4 P-21ESect-10.4 P-22ESect-10.4 P-23ESect-10.4 P-24ESect-10.4 P-25ESect-10.4 P-26ESect-10.4 P-27ESect-10.4 P-28ESect-10.4 P-29ESect-10.4 P-30ESect-10.4 P-31ESect-10.4 P-32ESect-10.4 P-33ESect-10.5 P-1CPSect-10.5 P-2CPSect-10.5 P-1ESect-10.5 P-2ESect-10.5 P-3ESect-10.5 P-4ESect-10.5 P-5ESect-10.5 P-6ESect-10.5 P-7ESect-10.5 P-8ESect-10.5 P-9ESect-10.5 P-10ESect-10.5 P-11ESect-10.5 P-12ESect-10.5 P-13ESect-10.5 P-14ESect-10.5 P-15ESect-10.5 P-16ESect-10.5 P-17ESect-10.5 P-18ESect-10.5 P-19ESect-10.5 P-20ESect-10.5 P-21ESect-10.5 P-22ESect-10.5 P-23ESect-10.5 P-24ESect-10.5 P-25ESect-10.5 P-26ESect-10.5 P-27ESect-10.5 P-28ESect-10.5 P-29ESect-10.5 P-30ESect-10.5 P-31ESect-10.5 P-32ESect-10.5 P-33ESect-10.5 P-34ESect-10.5 P-35ESect-10.5 P-36ESect-10.5 P-37ESect-10.5 P-38ESect-10.5 P-39ESect-10.5 P-40ESect-10.5 P-41ESect-10.5 P-42ESect-10.5 P-43ECh-10 P-1RECh-10 P-2RECh-10 P-3RECh-10 P-4RECh-10 P-5RECh-10 P-6RECh-10 P-7RECh-10 P-8RECh-10 P-9RECh-10 P-10RECh-10 P-11RECh-10 P-12RECh-10 P-13RECh-10 P-14RECh-10 P-15RECh-10 P-16RECh-10 P-17RECh-10 P-18RECh-10 P-19RECh-10 P-20RECh-10 P-21RECh-10 P-22RECh-10 P-23RECh-10 P-24RECh-10 P-25RECh-10 P-26RECh-10 P-27RECh-10 P-28RECh-10 P-29RECh-10 P-30RECh-10 P-31RECh-10 P-32RECh-10 P-33RECh-10 P-34RECh-10 P-35RECh-10 P-36RECh-10 P-37RECh-10 P-38RECh-10 P-39RECh-10 P-40RECh-10 P-41RECh-10 P-42RECh-10 P-43RECh-10 P-44RECh-10 P-45RECh-10 P-46RECh-10 P-47RECh-10 P-48RECh-10 P-49RECh-10 P-50RECh-10 P-51RECh-10 P-52RECh-10 P-53RECh-10 P-54RECh-10 P-55RECh-10 P-1TCh-10 P-2TCh-10 P-3TCh-10 P-4TCh-10 P-5TCh-10 P-6TCh-10 P-7TCh-10 P-8TCh-10 P-9TCh-10 P-10TCh-10 P-11TCh-10 P-12TCh-10 P-13TCh-10 P-14TCh-10 P-15TCh-10 P-16TCh-10 P-17T

Additional Math Solutions

Find more solutions based on key concepts

Show solutions add

Differentiate. f(t)=cottet

Single Variable Calculus: Early Transcendentals, Volume I

Simplify 4(3)25(2)2.

Intermediate Algebra

In Exercises 35-42, find functions f and g such that h = g f. (Note: The answer is not unique.) 42. h(x)=12x+1...

Applied Calculus for the Managerial, Life, and Social Sciences: A Brief Approach

Finding an Indefinite Integral In Exercises 35-56, find the indefinite integral. sinxcos3xdx

Calculus: Early Transcendental Functions (MindTap Course List)

Selecting midpoints of subintervals for xi to approximate the area under f(x) = 10x + 5 between x = 1 and x = 4...

Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th

The area of the shaded region is given by:

Study Guide for Stewart's Multivariable Calculus, 8th