Step by step
The question
Question: find and classify the critical points of f(x, y) = x³ − 12x − y² + 2y.
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.
12x gives 12.
y² has no x: it is a fixed number, so it gives 0.
2y has no x: it is a fixed number, so it gives 0.
Drop the zeros: ∂f/∂x = 3x² − 12.
Term by term.
x³ has no y: it is a fixed number, so it gives 0.
12x has no y: it is a fixed number, so it gives 0.
y²: the power 2 comes down: −1 times 2 is −2.
The power goes down by 1.
So 2y.
2y gives 2.
Drop the zeros: ∂f/∂y = −2y + 2.
Set ∂f/∂x = 0.
It has only x in it.
Move the number and divide by 3: x² = 4.
Two numbers square to 4: x = 2 or −2.
Set ∂f/∂y = 0.
It has only y in it.
Move the number: take away 2 on both sides.
Divide by −2: y = 1.
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.
12 has no x: it is a fixed number, so it gives 0.
Drop the zeros: f_xx = 6x.
Term by term.
2y gives 2.
2 has no y: it is a fixed number, so it gives 0.
Drop the zeros: f_yy = −2.
Term by term.
3x² has no y: it is a fixed number, so it gives 0.
12 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 (2, 1): f_xx is 12, f_yy is −2, f_xy is 0.
So D is −24.
Careful!
f_xx alone is not enough.
If D is negative, it is a saddle, whatever f_xx is.
D is negative: up one way, down the other.
A saddle point.
At (−2, 1): f_xx is −12, f_yy is −2, f_xy is 0.
So D is 24.
D is positive and f_xx is negative: a dome.
A local maximum.
There f is 17.
4 Check
Let's check.
Here is the surface near (2, 1).
It is a saddle: up one way, down the other.
Now small steps away from the point.
Near (2, 1), one step gives a bigger value, the other a smaller one.
That fits a saddle point.
Right!
Near (−2, 1), one step gives a smaller value, the other a smaller one.
That fits a local maximum.
Right!
Let's look back.
First: find the critical points.
Then: the second partials.
Then: test each point.
Last: check.