   Chapter 3.R, Problem 52E

Chapter
Section
Textbook Problem

# Use the guidelines in Section 3.5 to sketch the curve y = x sin x , 0 ≤ x ≤ 2 π . Use Newton’s method when necessary.

To determine

To sketch:

The curve of the given function

Explanation

1) Concept:

i. Find x-intercept and y-intercept and put x=0 of the given function.

ii. Symmetry: To find symmetry replace x by –x and check the behavior of function. Thus, if f-x=fx, then it is an even function, so it has y axis symmetry, and if f-x=-fx then it is an odd function, so it has x axis symmetry. And if f-x-fxf(x) then it has no symmetry.

iii. An asymptote is a tangent at infinity, to find horizontal, vertical and slant asymptote follow the rules.

iv. Function is increasing if f'x>0  and decreasing if f'x<0 in that particular interval.

v. The number f(c) is a local maximum value of f  if fcf(x) when x is near c. local minimum value of f if fc f(x) when x is near c.

vi. If f''x>0 function is concave up and f''x<0 function is concave down in that particular interval. And f''x=0 give the values of inflection points.

vii. Use Newton’s Method

2) Formula:

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

ii. Power rule of differentiation ddxxn=nxn-1

iii.

ddxconstant=0

3) Given:

y=xsinx,  0x2π

4) Calculation:

Here, the domain is 0, 2π

Intercepts:

To find y-intercept

Substitute x=0 in y, and solve it for y

y=0

y Intercept is (0, 0)

Now, to find x-intercept,

Substitute y=0 in given function and solve it for x.

0=xsinx

So, x-intercept are at x=0, π, 2π

Symmetry:

To find the axis of symmetry replace x by (-x)

Therefore,

f-x=-xsin-x

There is no symmetry.

But if fx is defined for all real numbers x then

f-x=xsinx

f(x) is an even function

Asymptote:

There is a polynomial function available in the given function, so therefore it has no asymptotes.

Intervals of increase or decrease:

To find the intervals of increase or decrease, we first find the derivative of the given function.

Differentiate f(x) with respect to x

f'x=xcosx+sinx

Again differentiate with respect to x

f''x=2cosx-xsinx

Draw the graphs of f'x & f''x

From the graph,

f'x  has solution at x2, 4.9

That means the critical numbers are x2, 4.9

To find these critical points more accurately,

Using Newton’s method for x2

xn+1=xn-f'xnf''xn

Set x1=2

x2=2-2cos2+sin22cos2-2sin2

x22.029048

For x3,

x3=2-2.029048cos2.029048+sin2.0290482cos2.029048-2.029048sin2.029048

x32.028758

For x4,

x4=2-2.028758cos2.028758+sin2.0287582cos2.028758-2.028758sin2.028758

x42.028758

Now, Using Newton’s method for x4.9

Set x1=4.9

x2=4.9-4.9cos4.9+sin4.92cos4.9-4.9sin4.9

x24.913214

For x3,

x3=4.913214-4.913214cos4.913214+sin4.9132142cos4.913214-4.913214sin4.913214

x34.913180

For x4,

x4=4.913180-4.913180cos4.913180+sin4.9131802cos4.913180-4.913180sin4.913180

x44.913180

So the critical numbers by using Newton’s method are

x=2.028758, 4.913180

By using these critical points and the domain, create three intervals as 0, 2.028758, 2.028758, 4.913180 & (4.913180, 2π)

Now, take a test point from each of the above intervals and check whether the function is increasing or decreasing in that interval.

For 0, 2.028758, consider x=1

f'1=1cos1+sin1

=0.99985+0.01745

=1.0173

f'x>0

The function is increasing in the interval 0, 2.028758

For 2.028758, 4.913180 , consider x=π

f'π=πcosπ+sinπ

=π-1+0

=-π

f'x<0

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