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 4.7, Problem 28E
To determine

To calculate: The absolute maximum value of curve,

  f(x)=xcosx,0xπ .

Expert Solution

Answer to Problem 28E

The maximum value with six decimal is x=0.561096 .

Explanation of Solution

Given information:

The curve is given as:

  f(x)=xcosx,0xπ .

Formula used:

Newton’s Method:

We seek a solution of f(x)=0 , starting from an initial estimate x=x1 .

For x=xn , compute the next approximation xn+1 by

  xn+1=xnf(xn)f'(xn) and so on.

Absolute Maximum Value: The critical numbers of function f within the interval [a,b] . Substitute the values in function and the largest value obtained by critical number is the absolute maximum value.

Calculation:

Consider the curve ,

  f(x)=xcosx,0xπ

Now,

  f(x)=xcosxf'(x)=x(sinx)+cosx

  =xsinx+cosx (by product rule)

Hence, we get the critical numbers from f'(x)=0 . In other words f'(x) is g(x) used to find out roots using Newton’s Method.

Therefore,

  g(x)=xsinx+cosx and

  g'(x)={x[cosx]+sinx}sinx

  =xcosxsinxsinx=xcosx2sinx

Now, let initial approximation be x1=0.8

For n=1

  x2=x1g(x1)g'(x1)

  x2=x1x1sinx1+cosx1x1cosx12sinx1

  x2=(0.8)(0.8)sin(0.8)+cos(0.8)(0.8)cos(0.8)2sin(0.8)x2=(0.8)0.573884872+0.6967067090.5573653671.434712182

  x2=0.80.122821837(1.992077549)x2=0.8+(0.061655148)x2=0.861655148

The second approximation is x2=0.861655148 .

Let x2=0.861655148

For n=2

  x3=x2g(x2)g'(x2)

  x3=x2x2sinx2+cosx2x2cosx22sinx2

  x3=(0.861655148){(0.861655148)sin(0.861655148)+cos(0.861655148)(0.861655148)cos(0.861655148)2sin(0.861655148)}x3=(0.861655148){0.653928535+0.6511822330.5610270171.51784281}

  x3=0.861655148{0.0027463022.078869827}x3=0.8616551480.001321055299x3=0.860334092

The third approximation is x3=0.860334092 .

The function values at critical numbers and endpoints are:-

  f(x)=xcosx

So, at end intervals

  f(0)=0cos(0)=0 ; and

  f(π)=πcos(π)=π(1)=π=3.141592654

At critical point x=0.860334092

  f(0.860334092)=0.860334092cos(0.860334092)=0.860334092(0.652184242)=0.561096338

Therefore, the absolute maximum value of curve f(x)=xcosx,0xπ is x=0.561096 .

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