Question

Use Newtons method to find the roots of the equation f(x)=x^3-3x-3 when x1=2

Answer 1

Could you solve the question and provide the answer

Answer 2

df/dx = 3 x^2 -3
X1= 2
then df/dx =12-3 = 9 =slope
f = 8-6-3= -1
so we go up from (2,-1) at slope 9
9 = change in f change in x = +1/change in x
change in x = 1/9
so
X2 = 2 1/9 = 19/9
etc

Answer 3

y = x^3 - 3x - 3
y' = 3 x^2 - 3
new x = old x + needed change in y / slope
X1 = 2, y = 8-6-3 = -1
so we need to go up 1
slope = 3(4)-3 = 9
---------------------
new x = 2 + 1/9
X2 = 19/9
start at X2 = 19/9 and do it again for the next guess, until you get there

Answer 4

In google paste:

Newton Raphson method calculator - AtoZmath.com

When page be open in rectangle type:

x^3-3x-3

Click option:

Initial solution x0 2

and click option:

Find

Four time click on option:

Click here to display next solution steps

You will see solution step by step.

Answer 5

I usually follow this algorithm:
f(x) = x^3-3x-3
f'(x) = 3x^2 - 3
Newton said: <b>newx = x - f(x)/f'(x)</b>
newx = x - (x^3 - 3x - 3)/(3x^2 - 3)
= (3x^3 - 3x - x^3 + 3x + 3)/(3x^2 - 3) = <b>(2x^3 + 3)/(3x^2 - 3)</b>

newx <--- x , start with x = 2
2.1111.... , keep replacing x with newx
2.1038359...
2.103803403
2.103803403 , correct to 9 decimals places