Question

Guys, I really need help with this problem. I need the solution within a half hour, if possible :)

Use Newton's method to approximate the positive root of sin(x) = x^3 correct to six decimal places.

Answer 1

I just got to 0.436383 (no idea if it's right) after starting with pi/6 (bad move), I'm fed up and u can do the rest yourself:-)

Answer 2

sin(x)-x^3=0
Let's guess x as 1
\[x_0-\frac{f\left(x_0\right)}{f'\left(x_0\right)}\]
plug in 1 in place of x
\[1-\frac{f(1)}{-3+\text{Cos}[1]}\]=0.935549
do another iteration

Answer 3

I guess taking a peek at Wolfram would have given a better guess:-)

Answer 4

Untruth! 45 minutes already, I didn't need to rush at all:-)