Math
posted by Jane .
Use Newton’s Method to approximate 3^(√7) to four decimal places.
Use x1 = 2 as your seed. Round off intermediate iterates to five decimal place

let f(x) = x^3  7
so f(∛7) = 0
See
http://keisan.casio.com/exec/system/1244946907
just enter your function, its derivative, and initial guess. 
the answer I should get is 18.2955
let x = 3^√7
then x^(1/√7) = 3
x^(1/√7)  3 = 0
let y = x^(1/√7)  3 = x^.3779645  3
dy/dx = (1/√7)^(1/√7  1) = .3779645 x^.62204
iteration expression:
x  (x^.3779645  3)/(.3779645x^.62204)
x = 2 > 15.06815
x = 15.06815 >18.10237..
x = 18.10237.. > 18.294858..
x = 18.29456 > 8.2955 , which was my calculator answer.
I find this a very strange question.
There is no practical way to do the above calculations without using a scientific calculator.
So why not find that scientific calculator to find the answer in the first place ? 
Good job, Reiny. I keep seeing 3^√ being used as cube root, so maybe I was way off.
In any case, the problem is solved.
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