   Chapter 3.8, Problem 34E

Chapter
Section
Textbook Problem

# Use Newton’s method to find the absolute maximum value of the function f ( x ) = x cos x , 0 ≤ x ≤ π , correct to six decimal places.

To determine

To find:

The absolute maximum value

Explanation

1) Concept:

i) Working rule of Newton’s method is,

Start with initial approximation x1  which is obtained from graph. Evaluate fx1and f '(x1) and apply the Newton’s formula. After the first iteration we get x2. Repeat the process with x1  replaced by x2. Calculate the approximations x1, x2, x3,  until the successive approximations xn, xn+1  agree the eight decimal places.

ii) Absolute Maximum value:

To find absolute maximum value of f find the zeros of f'(x) that is f'x=0, since f'x  exists for all x.

2) Formula:

i) Newton’s formula for nth approximation is xn+1=xn-fxnf'xn for n=1,2,3,

ii) Product Rule:

If f and g are both differentiable, then

ddxfxgx=fxddxgx+gxddx[fx]

3) Given:

fx=x cos x

4) Calculation:

Differentiating fx=x cos x  with respect to x,

By use of product rule,

ddxx cos x=xddxcos x+cos xddx[x]

Since, ddxcos x= -sinx and ddxx=1

f'(x)=x(-sinx)+cos x(1)

f'(x)=cosx-x sin x

We need to find zeros of f'(x), so find the next derivative of this function.

Again differentiating with respect to x,

f"(x)=ddx(cosx)-ddx(x sin x)

By use of product rule,

ddxx sin x=xddxsinx+sin xddx[x]

Since, ddxsinx=cosx and ddxx=1

=x(cosx)+sin x(1)

=xcosx+sin x

f"(x)=-sin x -(xcosx+sin x)

=-sinx-xcosx-sinx

=-2sinx-xcosx

Since we need zeros of f'x , get the graph of this function to examine the initial root,

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