BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 4, Problem 49RE
To determine

Tofind:theabsolute maximum value of the function using Newtonian method and correctup-to 8 decimal places.

Expert Solution

Answer to Problem 49RE

Theabsolute maximum value of the function is f(0.33541803)=1.16718557.

Explanation of Solution

Given:

  f(t)=cost+t-t2 .

Concept used:

If d2ydx2>0. the concave will be open upward and local minima can be found.

If d2ydx2<0. the concave will open downward and local maxima can be found.

Newton’s method formula:

  xn+1=xnf(xn)f(xn) .

Calculation:

The function is similar to a downward opening parabola so the local max will be the absolute maximum. Here need to find the zero of f .

  f(t)=cost+t-t2 .

  f(t)=-sint+1-2t

  lets consider the function interms of x .

  g(x)=sinx+12x .

  g(x)=cosx2 .

The graph of f(t)=cost+t-t2

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 4, Problem 49RE

Newton’s method formula:

  xn+1=xnf(xn)f(xn) .

  x1=0.3 .

  x2=0.3f(0.3)f(0.3)=0.3353529264 .

  x3=0.3353529264f(0.3353529264)f(0.3353529264)=0.3354180321 .

  x4=0.3354180321f(0.3354180321)f(0.3354180321)=0.3354180324 .

  f(t)=-cost-2 is negative for all t=0.33541803 .is where the absolute maximum is located.

The function value is

  f(0.33541803)=cos(0.33541803)+0.33541803-(0.33541803)2 .

  f(0.33541803)=1.16718557.

Hence the absolute maximum value of the function is f(0.33541803)=1.16718557.

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