Question

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Answer 1

☆ Finding the roots by Numerical Methods,

Given,

Solve, f(x) = xcos(x) - x, it's roots lies between (-4,4)

☆ We can plot the graph for visualising purpose with the help of matlab,

6 A 2 2 4 -6 2 2 LO OX (x)!

☆ The point where the function cuts the x-axis, that point is the root of the function.

☆ From the graph, we can see that the function f(x) cuts the x-axis in between (-4,4), more precisely we can say from graph is that, it cuts near to origin (0,0).

☆☆

☆ Without seeing graph, their is one more way to predict that function will have the roots in given interval or not.

☆ For a given interval, (a,b), If (f(a)×f(b)<0) then we can say that function will have root in given interval.

f(x) = xcos(x) - x,

f(a) = f(-4) = 6.6146,

f(b) = f(4) = -6.6146

f(a)f(b)<0,

As we can infer from above calculation is that the function have root in the interval (-4,4).

☆☆

Now, we will solve this for the roots by Newton's method,

Formula,

f(n f'(n n+1 n

For first iteration,

n = 0, x₀ = 1,

x_o is the initial guess, which is given in the question.

f(ro) f'(x)

f(x) = xcos(x) - x,

f'(x) = cos(x) - xsin(x) - 1,

1-0.459698 1=1-1.30117

10.6467041

2nd Iteration,

f(1) f'(x1)

(-0.130586) 20.6467041 (0.591603)

$\Rightarrow x_{2} = 0.425971$

3rd Iteration,

f(x2) a3 = 2f(x2)

$x_{3} = 0.425971 - \frac{(-0.0380656)}{(-0.265375)}$

3 = 0.282530

4th Iteration,

$x_{4} = x_3 - \frac{f(x_3)}{f'(x_3)}$

(-0.0112014) 40.282530 0.118412)

TA=0.187933

5th Interation,

$x_{5} = x_4 - \frac{f(x_4)}{f'(x_4)}$

$x_{5} = 0.187933 - \frac{(-0.003309)}{(-0.0527188)}$

6th iteration,

$x_{6} = x_5 - \frac{f(x_5)}{f'(x_5)}$

$x_{6} = 0.125166 - \frac{(-0.000979)}{(-0.0234487)}$

I6 = 0.08341

☆ Follow the same steps and it will take up closer to 0.

☆ Iteration can be done according to desired accuracy.

☆☆ Please comment below if you have any doubt regarding this question before rating this answer.