# Proving that a function has exactly four stationary points

I have the following function $f:\mathbb{R}^{2}\rightarrow \mathbb{R}$ defined by

$$f(x,y)=xy(x+y-1).$$

Using the property that the gradient in the stationary points is zero I calculated the four stationary points:

$$(0,0), (0,1), (1,0), (1/3,1/3)$$

in my homework assignment I now have to prove that the function has only these stationary points no more and no less. I don't really know how to do this can anyone help?

• You have to show that these are the only solutions of $\nabla f(x,y) = 0$. Sep 16, 2017 at 15:37

$$f_x(x,y) = 2xy + y^2 - y=y(2x+y-1)$$

$$f_y(x,y) = x^2 + 2xy+ x=x(x+2y+1)$$

Hence the stationary point must satisfy

($y=0$ or $2x+y-1=0$) and ($x=0$ or $x+2y+1=0$)

Consider each of the $4$ combinations, we are always have a linear system of which the system is non-singular (as the constraints are not parallel to each other), hence the solution for each system is unique. Hence there are at most $4$ solutions.

Since you have found $4$ solutions, there are exactly $4$ solutions.

from the first equation $$f_x=2xy+y^2-y$$ we get $$y(2x+y-1)=0$$ from here we get $$y=0$$ or $$y=1-2x$$ this can be plug into the second equation and this a poynomial $$5\,{x}^{2}-5\,x+1=0$$ and this has at least two Solutions.