Step by step
The question
Question: find and classify the critical points of f(x, y) = y³ − 3y + x².
1 Find the critical points
At a top, a bottom or a saddle, the surface is flat: both partial derivatives are 0.
Term by term.
y³ has no x: it is a fixed number, so it gives 0.
3y has no x: it is a fixed number, so it gives 0.
x²: the power 2 comes down.
The power goes down by 1.
So 2x.
Drop the zeros: ∂f/∂x = 2x.
Term by term.
y³: the power 3 comes down.
The power goes down by 1.
So 3y².
3y gives 3.
x² has no y: it is a fixed number, so it gives 0.
Drop the zeros: ∂f/∂y = 3y² − 3.
Set ∂f/∂x = 0.
It has only x in it.
Divide by 2: x = 0.
Set ∂f/∂y = 0.
It has only y in it.
Move the number and divide by 3: y² = 1.
Two numbers square to 1: y = 1 or −1.
Every x goes with every y: 2 critical points.
2 The second partials
Term by term.
2x gives 2.
So f_xx = 2.
Term by term.
3y²: the power 2 comes down: 3 times 2 is 6.
The power goes down by 1.
So 6y.
3 has no y: it is a fixed number, so it gives 0.
Drop the zeros: f_yy = 6y.
Term by term.
2x has no y: it is a fixed number, so it gives 0.
Drop the zeros: f_xy = 0.
The test number: D = f_xx × f_yy − (f_xy)².
3 Test each point
At (0, 1): f_xx is 2, f_yy is 6, f_xy is 0.
So D is 12.
Careful!
f_xx alone is not enough.
If D is negative, it is a saddle, whatever f_xx is.
D is positive and f_xx is positive: a bowl.
A local minimum.
There f is −2.
At (0, −1): f_xx is 2, f_yy is −6, f_xy is 0.
So D is −12.
D is negative: up one way, down the other.
A saddle point.
4 Check
Let's check.
Here is the surface near (0, 1).
It is a bowl: the point is at the bottom.
Now small steps away from the point.
Near (0, 1), one step gives a bigger value, the other a bigger one.
That fits a local minimum.
Right!
Near (0, −1), one step gives a bigger value, the other a smaller one.
That fits a saddle point.
Right!
Let's look back.
First: find the critical points.
Then: the second partials.
Then: test each point.
Last: check.