Chapter 4, Problem 50RE

To determine

Tofind:theroot of the Equation of the function using Newtonian method and correctup-to 6 decimal places.

Answer to Problem 50RE

Theabsolute maximum value of the function is $0.268881$ , $2.77005756$ .

Explanation of Solution

Given:

  $\sin x=x^2-3x+1$ .

Concept used:

If $\frac{{\text{d}}^2\text{y}}{{\text{dx}}^2}\text{>0}\text{.}$ the concave will be open upward and local minima can be found.

If $\frac{{\text{d}}^2\text{y}}{{\text{dx}}^2}\text{<0}\text{.}$ the concave will open downward and local maxima can be found.

Newton’s method formula:

  $x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}$ .

Calculation:

The function is similar to a downward opening parabola so the local max will be the absolute maximum. Here need to find the zero of $\text{<msup><mo>f</mo><mo>′</mo></msup>}$ ..

  $\text{lets consider the function interms of x}$ .

  $f\left(x\right)=\sin x-x^2+3x-1$ .

The graph of $\sin x=x^2-3x+1$

  

From the graph it can make easily approximate the initial value of x.

  $x\approx 0.3,2.8$ .

Starting with initial approximation $x_n$ .

  $x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}$ .

  $x_1=0.3$ .

  $x_2=0.3-\frac{f\left(0.3\right)}{f^{\prime }\left(0.3\right)}\approx 0.26855153$

  $x_3=0.26855153-\frac{f\left(0.26855153\right)}{f^{\prime }\left(0.26855153\right)}\approx 0.26888131$ .

  $x_4=0.26888131-\frac{f\left(0.26888131\right)}{f^{\prime }\left(0.26888131\right)}\approx \mathrm{0.268881.}$

The second root between 2 and 3.

  $x_1=2.8$

  $x_2=2.8-\frac{f\left(2.8\right)}{f^{\prime }\left(2.8\right)}\approx 2.77035425$ .

  $x_3=2.77035425-\frac{f\left(2.77035425\right)}{f^{\prime }\left(2.77035425\right)}\approx 2.77005759$ .

  $x_4=2.77005759-\frac{f\left(2.77005759\right)}{f^{\prime }\left(2.77005759\right)}\approx 2.77005756$ .

Hence the absolute maximum value of the function is $0.268881$ , $2.77005756$ .