newton raphson question

Question

Find a cut point of the $y = x^3-4x-5$ and $y = e^x-4x-5$ curves by selecting the starting point $x_0 = 3$, using the Newton Raphson method with an error of $10^{-3}$

I can't solve the question. Can you help me?

Answer 1

To calculate the cut point we set both function equal.

$x^3-4x-5= e^x-4x-5$

Next we subtract $e^x-4x-5$ on both side of the equation.

$$x^3-4x-5- e^x+4x+5=0\Rightarrow g(x)=x^3-e^x=0$$

So we want to evaluate the root of $g(x)$. The crucial formula of the Newton-Raphson method is

$$x_{n+1}=x_n-\dfrac{g(x_n)}{g'(x_n)}$$

Firstly we calculate the derivative $g'(x)=3x^2-e^x$. We have the initial value $x_0=3$. So you can go on and calculate $x_1$.

$$x_1=3-\frac{g(3)}{g'(3)}=3-\frac{3^3-e^3}{3\cdot 3^2-e^3}=2$$

The (absolute) error is $|x_1-x_0|=|2-3|=1$. This is greater than $10^{-3}$. So you have to go on.

$$x_2=2-\frac{g(2)}{g'(2)}=\ldots$$

            g(x)    x_(n)-x_(n-1) 
x1  2       0.61094 1
x2  1.8675  0.04092 0.13250
x3  1.8572  0.00025 0.01025
x4  1.8572  0.00000 0.00006