   Chapter 3.8, Problem 35E

Chapter
Section
Textbook Problem

# Use Newton’s method to find the coordinates of the inflection point of the curve y = x 2 sin x , 0 ≤ x ≤ π , correct to six decimal places.

To determine

To find:

Inflection points of the given curve

Explanation

1) Concept:

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.

Inflection point of y=fx  occur when y''=0 . So use the Newton’s method for g(x)=y''

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]

iii) Power Rule:

ddxxn=nxn-1

iv) Sum Rule:

If f and g are both differentiable, then

ddxfx+gx=ddxfx+ddx[gx]

3) Given:

y=x2sinx

4) Calculation:

Differentiating given function with respect to x,

By use of product rule,

ddxx2sinx=x2ddxsinx+sinxddx[x2]

By use of power rule, ddxx2=2x and ddxsinx=cosx

ddxx2sinx=x2(cosx)+sinx2x=x2cosx+2xsinx

f'(x)=x2cosx+2xsinx

Again differentiating with respect to x and by using sum rule,

ddxx2cosx+2xsinx=ddx(x2cosx)+ddx(2xsinx)

ddxx2cosx=x2ddxcosx+cosxddx[x2]

By use of power rule, ddxx2=2x and ddxcosx=-sinx

=x2(-sinx)+cosx(2x)

=-x2sinx+2xcosx

ddx2xsinx=2xddxsinx+sinxddx[2x]

=2x (cos x)+sinx(2)

=2x cos x+2sinx

f" (x)= -x2sinx+2xcosx+2x cos x+2sinx

f"(x) = -x2sinx+4xcosx+2sinx

Hence, f"(x) = 0 gives the inflection point in other words this is g(x) where we need to apply Newton’s method

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