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## Multivariable calculus

### Course: Multivariable calculus>Unit 3

Lesson 3: Optimizing multivariable functions

# Second partial derivative test example, part 1

A worked example of finding a classifying critical points of a two-variable function. Created by Grant Sanderson.

## Want to join the conversation?

• minor mistake you claim to have a grand total of three critical points while in fact you have 4
• I have the equation: x^2 + 6xy +10y^2 - 4y + 4. If I take the first derivative of x I end up with: 2x + 6y. What am I supposed to do with this? If I set it equal to zero I just keep ending up with x = -3y...?
• If you are looking for critical points, you will want to find the places where the tangent plane has zero slope. You will want to know where both partial df/dx and partial df/dy equal zero. In your example, you would calculate that partial df/dy is 6x +20y-4. Now you have two equations equal to zero with two variables. Just use algebra to solve the system of equations.
• What about inflection points? Where do they fit in in the gradient and the second partial derivative test?
(1 vote)
• but what about critical point from the partial derivative with respect to x (0,1) why don't we also consider it being a critical point in the final answer? Many thanks.
(1 vote)
• If you go to Desmos and you graph the two equations mentioned in this video, https://www.desmos.com/calculator/d5nrguizbn
you'll see that one of them just boils down to the horizontal line y=1 (and it's undefined at x=0) while the other one is just some hyperbola. As far as I can tell, there are only 2 intersections between the graphs, so although it made sense from the video why there should be 4 critical points, how does it square with this graphical view of the situation?
(1 vote)
• However, it does seem valid according to this 3d Geogebra graph for the full multivariable function f(x,y): https://www.geogebra.org/3d/fq3dutqv
(1 vote)
• Can you try not to make sound of swallowing saliva when you are talking. It's disturbing.