Find the third approximation of x3 of Newton's method to approximate the intersection of $\displaystyle y=[\sqrt(x)]+1$ and $\displaystyle y=x^2$.
What do I do? Set them equal to each other?
So I did $\displaystyle f(x)=x^2-\sqrt(x)-1$
$\displaystyle f'(x)= 2x-\frac{1}{2\sqrt(x)}$
I used $\displaystyle x1=1$
$\displaystyle x2= 1 + \frac{1-1-1}{2-.5}$
= .333
$\displaystyle x3=.333 + \frac{-1.466}{-.8660}$
=$\displaystyle 2.0227$
Answer is supposed to be: $\displaystyle 1.50143$