Question

Hi, I have a difficult calculus type question, ive been working on it for ages, Need Help

Answer 1

Attached is the question

Answer 2

someone else had this very same question earlier, let me see if i can find the post

Answer 3

\[
L(x) = \sqrt{(0.2-x)^2+(2.7-y)^2} = \sqrt{(0.2-x)^2+(2.7-x^2)^2}
\]

Answer 4

if it was me, i would get rid of that nasty square root, and work with the square of the distance instead
it will make finding the derivative and differentiation tons easier

Answer 5

sorry i cant find it

Answer 6

anyway once you differentiate and set numerator equal to 0
here is how you do Newtons Method:
\[\large x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)}\]
f(x) is numerator of L'(x)
x_0 is initial guess ... i would try x=1

you keep repeating the process until x_n+1 = x_n or f(x) = 0
it usually takes about 5 iterations...plugging formula into Excell will make it faster

Answer 7

you should end up with
\[f(x) = 4x^{3}-8.8x-0.4 = 0\]
\[\rightarrow f(x) =x^{3}-2.2x-0.1 = 0\]
\[f'(x) = 3x^{2}-2.2\]

Answer 8

@dumbcow I got the f(x) and f'(x), im trying to find out the newton's method

Answer 9

And then when i have to find the global minimum, do i just use the equation equal it to zero and second derivative? or is it also through newtons?

Answer 10

well i posted Newtons method...which will give you the 3 zeros
**Note** the zero you obtain depends on the initial guess, make sure they are far apart

as far as global min, its just the smallest value of L(x) after you plug in the 3 zeros obtained from Newtons method

Answer 11

Oh thanks

Answer 12

also the diagram gives you a hint, obviously the point is somewhere around x=1.5
so the global min is value from zero closest to 1.5

Answer 13

I posted this solution yesterday

Answer 14

@CarlosGP , very nicely done thanks for posting it again

Answer 15

My pleasure!

Answer 16

Here is yesterday's link. The thread can be useful for further clarification
http://openstudy.com/users/carlosgp#/updates/523f95dfe4b0900154177a8c