Chapter 4.7, Problem 28E

To determine

To calculate: The absolute maximum value of curve,

  $f(x)=x\cos x,0\le x\le \pi$ .

Answer to Problem 28E

The maximum value with six decimal is $x=0.561096$ .

Explanation of Solution

Given information:

The curve is given as:

  $f(x)=x\cos x,0\le x\le \pi$ .

Formula used:

Newton’s Method:

We seek a solution of $f(x)=0$ , starting from an initial estimate $x=x_1$ .

For $x=x_n$ , compute the next approximation $x_{n+1}$ by

  $x_{n+1}=x_n-\frac{f(x_n)}{f\text{'}(x_n)}$ and so on.

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

Calculation:

Consider the curve ,

  $f(x)=x\cos x,0\le x\le \pi$

Now,

  $\begin{array}{l}f(x)=x\cos x\\ f\text{'}(x)=x(-\sin x)+\cos x\end{array}$

  $=-x\sin x+\cos x$ (by product rule)

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

Therefore,

  $g(x)=-x\sin x+\cos x$ and

  $g\text{'}(x)=-\left\{x\left[\cos x\right]+\sin x\right\}-\sin x$

  $\begin{array}{l}=-x\cos x-\sin x-\sin x\\ =-x\cos x-2\sin x\end{array}$

Now, let initial approximation be $x_1=0.8$

For $n=1$

  $x_2=x_1-\frac{g(x_1)}{g\text{'}(x_1)}$

  $x_2=x_1-\frac{-x_1\sin x_1+\cos x_1}{-x_1\cos x_1-2\sin x_1}$

  $\begin{array}{l}{x}_2=(0.8)-\frac{-(0.8)\sin (0.8)+\cos (0.8)}{-(0.8)\cos (0.8)-2\sin (0.8)}\\ x_2=(0.8)-\frac{-0.573884872+0.696706709}{-0.557365367-1.434712182}\end{array}$

  $\begin{array}{l}{x}_2=0.8-\frac{0.122821837}{(-1.992077549)}\\ x_2=0.8+(0.061655148)\\ x_2=0.861655148\end{array}$

The second approximation is $x_2=0.861655148$ .

Let $x_2=0.861655148$

For $n=2$

  $x_3=x_2-\frac{g(x_2)}{g\text{'}(x_2)}$

  $x_3=x_2-\frac{-x_2\sin x_2+\cos x_2}{-x_2\cos x_2-2\sin x_2}$

  $\begin{array}{l}{x}_3=(0.861655148)-\left\{\frac{-(0.861655148)\sin (0.861655148)+\cos (0.861655148)}{-(0.861655148)\cos (0.861655148)-2\sin (0.861655148)}\right\}\\ x_3=(0.861655148)-\left\{\frac{-0.653928535+0.651182233}{-0.561027017-1.51784281}\right\}\end{array}$

  $\begin{array}{l}{x}_3=0.861655148-\left\{\frac{-0.002746302}{-2.078869827}\right\}\\ x_3=0.861655148-0.001321055299\\ x_3=0.860334092\end{array}$

The third approximation is $x_3=0.860334092$ .

The function values at critical numbers and endpoints are:-

  $f(x)=x\cos x$

So, at end intervals

  $\begin{array}{l}f(0)=0\cos (0)\\ =0\end{array}$ ; and

  $\begin{array}{l}f(\pi )=\pi \cos (\pi )\\ =\pi (-1)\\ =-\pi \\ =-3.141592654\end{array}$

At critical point $x=0.860334092$

  $\begin{array}{l}f(0.860334092)=0.860334092\cos (0.860334092)\\ =0.860334092(0.652184242)\\ =0.561096338\end{array}$

Therefore, the absolute maximum value of curve $f(x)=x\cos x,0\le x\le \pi$ is $x=0.561096$ .