
Newtons Method
I am working on problems using Newton's Method in which i'm given an equation and am told to find a root between 2 numbers.
For this particular question there is no 'x' in the equation so I'm not quite sure what to do:
Estimate $\displaystyle ^4\sqrt{19}$ to 2 decimal place accuracy. Use 2 as your initial estimate.

Make $\displaystyle f(x) = x^{1\over 4}$

could you explain further please

I understand Newton's Method to be
$\displaystyle x_{n+1} = x_{n}\frac{f(x_{n})}{f'(x_{n})}$
Now find $\displaystyle f'(x)$ and start iterating.

So I'm using $\displaystyle f(x) = x^\frac{1}{4}$
and the derivative $\displaystyle f'(x) = \frac{1}{4}x^{\frac{3}{4}}$ and iterating until it equals $\displaystyle ^4\sqrt{19}$ ?

No!
Use $\displaystyle f(x)=x^4.$
Now you are looking for the root of $\displaystyle x^419=0$
$\displaystyle f'(x)=4x^3$
Take a first shot at the answer, call it $\displaystyle x_n$
Say $\displaystyle x_n=4$
Then $\displaystyle x_{n+1}=x_n\frac{f(x_n)}{f'(x_n)}$

I just want to clean this up,
as i answered in a very general way...
$\displaystyle f(x)=x^419$
$\displaystyle f'(x)=4x^3$
This is a very steep curve and an initial guess at the solution
can be within a range of x.
Simplest is to take the square root of the square root of 16,
or the fourth root of 16 which is 2.
$\displaystyle x_n=2$
The next approximation to the 4th root of 19 is then found from
$\displaystyle x_{n+1}=x_n\frac{f(x_n)}{f'(x_n)}=2\frac{f(2)}{f'(2)}$
$\displaystyle x_{n+1}=2\frac{2^419}{(4)2^3}=2\frac{3}{32}$
Note... $\displaystyle (2+\frac{3}{32})^4=19.2176$