Step by step
The question
Question: find and classify the critical points of f(x, y) = x³ − 3x + y².
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.
x³: the power 3 comes down.
The power goes down by 1.
So 3x².
3x gives 3.
y² has no x: it is a fixed number, so it gives 0.
Drop the zeros: ∂f/∂x = 3x² − 3.
Term by term.
x³ has no y: it is a fixed number, so it gives 0.
3x has no y: it is a fixed number, so it gives 0.
y²: the power 2 comes down.
The power goes down by 1.
So 2y.
Drop the zeros: ∂f/∂y = 2y.
Set ∂f/∂x = 0.
It has only x in it.
Move the number and divide by 3: x² = 1.
Two numbers square to 1: x = 1 or −1.
Set ∂f/∂y = 0.
It has only y in it.
Divide by 2: y = 0.
Every x goes with every y: 2 critical points.
2 The second partials
Term by term.
3x²: the power 2 comes down: 3 times 2 is 6.
The power goes down by 1.
So 6x.
3 has no x: it is a fixed number, so it gives 0.
Drop the zeros: f_xx = 6x.
Term by term.
2y gives 2.
So f_yy = 2.
Term by term.
3x² has no y: it is a fixed number, so it gives 0.
3 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 (1, 0): f_xx is 6, f_yy is 2, 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 (−1, 0): f_xx is −6, f_yy is 2, 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 (1, 0).
It is a bowl: the point is at the bottom.
Now small steps away from the point.
Near (1, 0), one step gives a bigger value, the other a bigger one.
That fits a local minimum.
Right!
Near (−1, 0), 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.