How to find any absolute extrema on $\mathbb R^2$ for $y^4 - x^5$

Question

So far I have found the critical point $(0, 0)$ and the partials of $-5x^4$ and $4y^3$. How would I proceed after finding the critical point, since I don't have a definite interval. I get confused without an interval.

Answer 1

The critical points of $f(x,y)=y^4-x^5$ are:

• All the points in the border
• All the points with $\nabla f=(0,0)$

Since there is no border, the critical points are just the last indicated, so: $$ \nabla f=(-5x^4,4y^3)=(0,0) \implies (x,y)=(0,0) $$

Hence the function has only one critical point, which turns to be only a saddle point (why?).