Answer
$x_3 = 1.442250719$
Work Step by Step
$f(x) = x^3 - 3$ and $x_1 = 1.4 = \dfrac{7}{5}$
I will complete this method in two separate ways, using one to calculate $x_2$ and one to calculate $x_3$ (Either method works). The method for $x_2$ should be used if you DO NOT HAVE a CALCULATOR and $x_3$ should be used if you DO have a calculator. Find a formula to approximate the zero of $f(x)$.
METHOD 1.) Using the textbook's equation: $x_{n + 1} = x_n - \dfrac{f(x)}{f'(x)}$ come up with a formula to use. Just plug in $f(x)$ and $f'(x)$.
$x_{n+1} = x - \dfrac{x^3 - 3}{3x^2} = \dfrac{3x^3 - x^3 + 3}{3x^2} = \dfrac{2x^3 + 3}{3x^2} \rightarrow$ This is our formula.
Plug in 1.4 into x in the formula to compute $x_2$
$x_2 = \dfrac{2(\dfrac{7}{5})^3 + 3}{3(\dfrac{7}{5})^2} = \dfrac{2(\dfrac{343}{125}) + 3}{3(\dfrac{49}{25})} = \dfrac{\dfrac{686}{125} + \dfrac{375}{125}}{\dfrac{147}{25}} = \dfrac{\dfrac{1061}{125}}{\dfrac{147}{25}} = \dfrac{1061}{5(147)} = \dfrac{1061}{735}$
You can see how everything was kept in a fraction. This method is obviously a pain but useful if you don't have a calculator.
METHOD 2.) Use $x_{n + 1} = x_n - \dfrac{f(x)}{f'(x)}$ to compute $x_3$
$x_{3} = x_2 - \dfrac{f(x_2)}{f'(x_2)} = \dfrac{1061}{735} - \dfrac{f(\dfrac{1061}{735})}{f'(\dfrac{1061}{735})} = 1.442250719$
