# Tofind: theabsolute maximum value of the function using Newtonian method and correctup-to 8 decimal places. ### Single Variable Calculus: Concepts...

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

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 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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