Critical point of $f(x,y) = x^4 - 4x^3 + 2y^2 + 8xy + 1$

Question

I’m solving for the critical points of $f(x,y) = x^4 - 4x^3 + 2y^2 + 8xy + 1$ by finding the partial derivatives wrt x and y.

$$f_x(x,y) = 4x^3 - 12x^2 + 8y$$ $$f_y(x,y) = 4y - 8x$$

after getting the partial derivatives, equate them to 0 to get the values of x or y by factoring. With that, I used $f_y$ so I have

$$4(y-2x)=0$$ $$y=2x$$

Then I used that in substituting to the equation of $f_x$, so I have

$$4x^3 - 12x^2 + 8(2x) =0 $$ $$4x^3 - 12x^2 + 16x =0$$ $$4x(x^2-3x+4)=0$$

So I got is $x=0$ then substitute to $y=2x$ which makes $y=0$. And the critical point is (0,0).

But when I tried online calculators, there’s other 2 critical points but I can’t seem to find the other points. The 2 other points were (-1,2) and (4,-8).

Answer 1

Carefully check that $$f_y(x,y) = 4y + 8x$$ and hence $y=-2x$.

Using this, the next equation is $4x(x^2-3x\color{red}{-}4=0)=0$ which gives both $x=0$ and $x^2-3x\color{red}{-}4=0\implies x=-1,4$ and hence the other two solutions.