×
Log in to StudySoup
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4 - Problem 15re
Join StudySoup for FREE
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4 - Problem 15re

Already have an account? Login here
×
Reset your password

Curve sketching Use the guidelines of this | Ch 4 - 15RE

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett ISBN: 9780321570567 2

Solution for problem 15RE Chapter 4

Calculus: Early Transcendentals | 1st Edition

  • Textbook Solutions
  • 2901 Step-by-step solutions solved by professors and subject experts
  • Get 24/7 help from StudySoup virtual teaching assistants
Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Calculus: Early Transcendentals | 1st Edition

4 5 1 270 Reviews
13
4
Problem 15RE

Curve sketching? Use the guidelines of this chapter to make a complete graph of the following functions on their domains or on the given interval. Use a graphing utility to check your work. f(x) = 4 cos[?(x?1)]on [0,2]

Step-by-Step Solution:

Solution Step 1 In this problem we need to make a complete graph of f(x) = 4 cos [(x1)]in the given interval [0,2] In order to sketch the complete graph, we need to find the critical points, inflection points, local maximum and local minimum if possible. First let us see the definitions: Critical point: An interior point cof the domain of a function f at which f (c) = 0or f(c)fails to exist is called a critical point of f Inflection Point: An inflection point is a point on a curve at which the sign of the curvature (i.e., the concavity) changes. Inflection points may be stationary points, but are not local maxima or local minima. A necessary condition for x to be an inflection point is f (x) = 0 Local maximum: Let f be function defined on an interval [a,b]and let pbe a point in the open interval (a,b). Then the function f has local maximum at pif f(p) f(x) for all xin the neighborhood of the point p. Local minimum: Let f be function defined on an interval [a,b]and let pbe a point in the open interval (a,b). Then the function f has local minimum at pif f(p) f(x)for all x in the neighborhood of the point p. Step 2 Consider the following function The given interval is [0,2]. First put x = 0into the function, we get Now put x = 0.5 into the function, we get Similarly, evaluate the functional values at some more points The table showing the function values is shown below x f(x) 0 -4 0.5 0 1 4 1.5 0 2 -4 Therefore, the graph of the function is shown below Step 3 Now, at the points (0,-4) and (2,-4), the plot has the lowest value of the function. Therefore by the definition in step 1, these two points are the lowest point for the entire plot. So, this will be absolute minima Also consider the points (0.5,0) and (1.5,0). Note that the concavity of the graph changes at both of these points. Therefore by the definition in step 1, these two points are the inflection points Also, note that the graph of the function attains its maximum value at (1,4), Therefore by the definition in tep 1, the absolute maxima of the graph is at (1,4) So, the complete graph of the function f(x) = 4 cos[(x1)]is

Step 4 of 6

Chapter 4, Problem 15RE is Solved
Step 5 of 6

Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

Other solutions

People also purchased

Related chapters

Unlock Textbook Solution

Enter your email below to unlock your verified solution to:

Curve sketching Use the guidelines of this | Ch 4 - 15RE