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 50RE
To determine

Tofind:theroot of the Equation of the function using Newtonian method and correctup-to 6 decimal places.

Expert Solution

Answer to Problem 50RE

Theabsolute maximum value of the function is 0.268881 , 2.77005756 .

Explanation of Solution

Given:

  sinx=x23x+1 .

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 ..

  lets consider the function interms of x .

  f(x)=sinxx2+3x1 .

The graph of sinx=x23x+1

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 4, Problem 50RE

From the graph it can make easily approximate the initial value of x.

  x0.3,2.8 .

Starting with initial approximation xn .

  xn+1=xnf(xn)f(xn) .

  x1=0.3 .

  x2=0.3f(0.3)f(0.3)0.26855153

  x3=0.26855153f(0.26855153)f(0.26855153)0.26888131 .

  x4=0.26888131f(0.26888131)f(0.26888131)0.268881.

The second root between 2 and 3.

  x1=2.8

  x2=2.8f(2.8)f(2.8)2.77035425 .

  x3=2.77035425f(2.77035425)f(2.77035425)2.77005759 .

  x4=2.77005759f(2.77005759)f(2.77005759)2.77005756 .

Hence the absolute maximum value of the function is 0.268881 , 2.77005756 .

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