Question

Use the guidelines in section 4.4 to sketch the curve\({\bf{y = xsinx,0}} \le {\bf{x}} \le {\bf{2\pi }}\) . Use Newton’s method when necessary.

Short answer

A) Doman: \(0 \le x \le 2\pi \)

B) \(x\) intercepts at \(x = 0,\pi ,2\pi \) and the \(y\) intercept is 0.

C) No symmetry

D )No Asymptotic

The graph of the function is:

Step-by-step solution

Find domain and intercept of the given function.

The given function is \(y = x\sin x\).

A.Domain: \(0 \le x \le 2\pi \)

B.Intercepts of the function:

Let \(x = 0\) then it gives,

\(\begin{aligned}{c}y &= 0\sin 0\\ &= 0\end{aligned}\)

So, \(y\) intercept only at \(y = 0\).

Now, put \(y = 0\) in the given function gives,

\(0 = x\sin x\)

Thus, true when \(x = 0\) or \(\sin x = 0\).

So, \(x\) intercepts at \(x = 0,\pi ,2\pi \).

Symmetry and Asymptotes of the given function

C.Symmetry:

No symmetry within \(0 \le x \le 2\pi \).

D. Asymptotic:

No asymptotic because \(x\) is defined foe all values in the interval.

Intervals of increasing or decreasing of the given function

E. Intervals of function for increase or decrease:

The derivative of the given function is as follows:

\(\begin{aligned}{l}f\left( x \right) = x\sin x\\f'\left( x \right) = x\cos x + \sin x\\f''\left( x \right) = 2\cos x - x\sin x\end{aligned}\)

Now, use Newton’s method to calculate the \(x\) intercepts of the derivative.

\({x_{n + 1}} = {x_n} - \frac{{f'\left( {{x_n}} \right)}}{{f''\left( {{x_n}} \right)}}\)

Take \({x_1} = 2\) then it gives,

\(\begin{aligned}{c}{x_2} &= 2 - \frac{{2\cos 2 + \sin 2}}{{2\cos 2 - 2\sin 2}}\\ &= 2.029048\end{aligned}\)

Similarly, Take \({x_2} = 2.029048\) then it gives,

\(\begin{aligned}{c}{x_3} &= 2.029048 - \frac{{2.029048\cos 2.029048 + \sin 2.029048}}{{2.029048\cos 2.029048 - 2.029048\sin 2.029048}}\\ &= 2.028758\end{aligned}\)

Thus, one intercept is at \(x \approx 2.03\).

Then, take \({x_1} = 5\)

\(\begin{aligned}{c}{x_2} &= 5 - \frac{{5\cos 5 + \sin 5}}{{5\cos 5 - 5\sin 5}}\\ &= 4.914325\end{aligned}\)

Similarly, solve as follows:

\(\begin{aligned}{c}{x_3} &= 4.914325 - \frac{{4.914325\cos 4.914325 + \sin 4.914325}}{{4.914325\cos 4.914325 - 4.914325\sin 4.914325}}\\ &= 4.913181\end{aligned}\)

Thus, another intercept is at \(x \approx 4.91\).

Finally, to find the intervals of increase or decrease by putting in values to \(f'\left( x \right)\) with each interval.

\(\begin{aligned}{c}f'\left( 1 \right) &= 1\cos 1 + \sin 1\\ &= 1.38\end{aligned}\)

Thus, \(f'\left( x \right)\) is positive then \(f\left( x \right)\) is increasing on the interval \(0 < x < 2.03\)

\(\begin{aligned}{c}f'\left( 3 \right) = 3\cos 3 + \sin 3\\ = - 2.38\end{aligned}\)

Thus, \(f'\left( x \right)\) is negative then \(f\left( x \right)\) is decreasing on the interval \(2.03 < x < 4.91\)

\(\begin{aligned}{c}f'\left( 5 \right) &= 5\cos 5 + \sin 5\\ &= 0.46\end{aligned}\)

Thus, \(f'\left( x \right)\) is positive then \(f\left( x \right)\) is increasing on the interval \(4.91 < x < 2\pi \)

Local maximum or minimum values of the given function

F.Local maximum and minimum values:

The extreme values of the given function are \(x = 2.03,4.91\).

On the basis of the intervals of increase and decrease the maximum point is at \(x = 2.03\) because the function increases before and decreases after.

Similarly, \(x = 4.91\) is a minimum because the function decreases before and increases after.

Concavity and Points of inflection of the given function

G. Concavity and Points of inflection:

As obtained that \(f''\left( x \right) = 2\cos x - x\sin x\) then, to find the points of inflection solve as follows:

\(\begin{aligned}{c}2\cos x - x\sin x &= 0\\2\cos x &= x\sin x\end{aligned}\)

Now, use Newton’s method again and find the intercepts of \(f''\left( x \right)\) as follows:

\({x_{n + 1}} = {x_n} - \frac{{f''\left( {{x_n}} \right)}}{{f'''\left( {{x_n}} \right)}}\)

Differentiate \(f''\left( x \right)\) and it gives,

\(f'''\left( x \right) = - 3\sin x - x\cos x\)

Take \({x_1} = 1\) then it gives,

\(\begin{aligned}{c}{x_2} &= 1 - \frac{{1\cos 1 - 1\sin 1}}{{ - 3\sin 1 - 1\cos 1}}\\ &= 1.078028\end{aligned}\)

Similarly, Take \({x_2} = 1.078028\) then it gives,

\(\begin{aligned}{c}{x_3} &= 1.078028 - \frac{{2\cos 1.078028 - 1.078028\sin 1.078028}}{{ - 3\sin 1.078028 - 1.078028\cos 1.078028}}\\ &= 1.076874\end{aligned}\)

Thus, one intercept is at \(x \approx 1.077\).

Then, take \({x_1} = 4\)

\(\begin{aligned}{c}{x_2} &= 4 - \frac{{2\cos 4 - 4\sin 4}}{{ - 3\sin 4 - 4\cos 4}}\\ &= 3.647916\end{aligned}\)

Similarly, solve as follows:

\(\begin{aligned}{c}{x_3} = 3.647916 - \frac{{3.647916\cos 3.647916 - 3.647916\sin 3.647916}}{{ - 3\sin 3.647916 - 3.647916\cos 3.647916}}\\ = 3.647916\end{aligned}\)

Thus, another intercept is at \(x \approx 3.644\).

Finally, to find the concavity by putting in values to \(f''\left( x \right)\) with each interval.

\(\begin{aligned}{c}f''\left( 1 \right) &= 2\cos 1 - 1\sin 1\\ &= 0.239\end{aligned}\)

Thus, \(f''\left( x \right)\) is positive then \(f\left( x \right)\) is concave up on the interval \(0 < x < 1.077\)

\(\begin{aligned}{c}f''\left( 2 \right) &= 2\cos 2 - 2\sin 2\\ &= - 2.651\end{aligned}\)

Thus, \(f''\left( x \right)\) is negative then \(f\left( x \right)\) is concave down on the interval \(1.077 < x < 3.644\)

\(\begin{aligned}{c}f''\left( 5 \right) &= 2\cos 5 - 5\sin 5\\ &= 5.362\end{aligned}\)

Thus, \(f'\left( x \right)\) is positive then \(f\left( x \right)\) is concave up on the interval \(3.644 < x < 2\pi \)

Graph of the given function.

The graph of the function is as follows:

Hence, A) Doman: \(0 \le x \le 2\pi \)

B) \(x\) intercepts at \(x = 0,\pi ,2\pi \) and the \(y\) intercept is 0.

C)No symmetry

D)No Asymptotic