# Nonlinear programming

• Dec 12th 2010, 11:20 AM
Keep
Nonlinear programming
I am trying to do a non-linear programming work. We have two functions: $f(x)=x^2/2+e^{-x}$ and $g(x) = x^4-32x^2$. I just want to know how exactly one can get the minimum of x, x*. For example using the newton method:

Code:

```k    ak          F'(x)    F''(X)      k    ak            g'(x)    g''(x) 1    2,0000    1,8647    1,1353        1    0,2000    -12,7680    -63,5200 2    0,3576    -0,3417    1,6993        2    -0,0010    0,0645    -64,0000 3    0,5587    -0,0132    1,5719        3    0,0000    0,0000    -64,0000 4    0,5671    0,0000    1,5672                    5    0,5671    0,0000    1,5671```
So I caluclate and get this but then how can I determine the minimum of x, x*?
• Dec 12th 2010, 12:17 PM
pickslides
Well the minimum for the function f(x) will be where f'(x)= 0 and f''(x*)>0.

Same goes for g(x)
• Dec 12th 2010, 01:25 PM
Keep
OK I understand f'(x) but what did you mean by f"(x*) or you meant f"(x)? Because I am looking for the minimum, x*, so how can I find f"(x*) when I don't even have x*? If you mean the minimum, x*, is where f'(x) = 0 and f"(x)>0, then yeah I understand. So that would be x* = 0.5671 for f(x) but then I don't know how I can get x* for g(x) because although we have a place where f'(x) = 0, we have no where where f"(x)>0?

• Dec 12th 2010, 02:49 PM
Keep
OK I read a some stuff on the internet and I found out that the minimum is where f'(x) = 0 and f''(x)>0. If f''(x)<0, then that is a maximum point but in the question we were asked to find out the minimum, not the maximum so I am confused. When f'(x) = 0, f"(x)<0 so I am wondering how I can find the min. for this equation.
• Dec 12th 2010, 02:54 PM
pickslides
Quote:

Originally Posted by Keep
OK I understand f'(x) but what did you mean by f"(x*) or you meant f"(x)? Because I am looking for the minimum, x*, so how can I find f"(x*) when I don't even have x*? If you mean the minimum, x*, is where f'(x) = 0 and f"(x)>0, then yeah I understand. So that would be x* = 0.5671 for f(x) but then I don't know how I can get x* for g(x) because although we have a place where f'(x) = 0, we have no where where f"(x)>0?

x* is the solution to f'(x)=0.
• Dec 12th 2010, 02:59 PM
pickslides
Quote:

Originally Posted by Keep
OK I read a some stuff on the internet and I found out that the minimum is where f'(x) = 0 and f''(x)>0. If f''(x)<0, then that is a maximum point but in the question we were asked to find out the minimum, not the maximum so I am confused. When f'(x) = 0, f"(x)<0 so I am wondering how I can find the min. for this equation.

$g(x) = x^4-32x^2$

$g'(x)= 4x^3-64x$

making $g'(x)=0$ to solve

$0 = 4x^3-64x$

$0 = 4x(x^2-16) = 4x(x-4)(x+4)$

$x^* = 0,-4,4$

Now find $g(x^*)$
• Dec 12th 2010, 05:48 PM
Keep
OK thanks. Substituting we get: for x* = 0, g(x*) = 0, for x* = +/-4, g(x)= -256. So out of those three values for x* which one is the minimum?