# Finding the critical points of a multivariable function

• Koi9
In summary, the function has two critical points at x = ±√7 and three critical points at y = 0, ±√3. However, since the Second Derivative Test requires fy to be nonzero at the critical point, only the solutions at y = ±√3 can be classified. The critical points at x = ±√7 are classified as "undetermined" due to the test failing.

## Homework Statement

\
Find the critical points of the function. Then use the Second Derivative Test to classify each critical point (or state that the test fails).

f(x, y) = x3 + y4 - 21x - 18y2

## The Attempt at a Solution

partial x derivative=3x^2-21=0

partial y= 4y^3-36y=0

i try to solve those for numbers, but the answer space in the question has space for 6 different critical point answers, so I am not sure where to go from here.

Koi9 said:

## Homework Statement

\
Find the critical points of the function. Then use the Second Derivative Test to classify each critical point (or state that the test fails).

f(x, y) = x3 + y4 - 21x - 18y2

## The Attempt at a Solution

partial x derivative=3x^2-21=0

partial y= 4y^3-36y=0

i try to solve those for numbers, but the answer space in the question has space for 6 different critical point answers, so I am not sure where to go from here.
What did you get when you solved these for numbers?

You should get two solutions for x and three solutions for y.

SammyS said:
What did you get when you solved these for numbers?

You should get two solutions for x and three solutions for y.

partial x = 3x^2-21=0
3x^2=21
x=sqrt(21)/3

partial y = 4y^3-36y=0
4y^3=36y
4y^2=36
= 6
y=3/2

must be missing something, because I don't really see another way to solve it.Thanks,
Matt

Last edited:
Assignment due soon, anyone help please?

Koi9 said:
partial x = 3x^2-21=0
3x^2=21
x=sqrt(21)/3

partial y = 4y^3-36y=0
4y^3=36y
4y^2=36
= 6
y=3/2

must be missing something, because I don't really see another way to solve it.

Thanks,
Matt

That should be $x=\pm\sqrt{\frac{21}{3}}=\pm\sqrt{7}\,.$

Similarly, use ± for the y solution.

Also, when you divided by y, you got rid of the y=0 solution.

Koi9 said:
partial x = 3x^2-21=0
3x^2=21
x=sqrt(21)/3

partial y = 4y^3-36y=0
4y^3=36y
4y^2=36
= 6
y=3/2
In addition to what SammyS said, there's another error here.
Setting fy = 0 yields
4y3-36y=0
==> 4y(y2 - 9) = 0
I hope you can see that this equation has three solutions, none of which is 3/2.
Koi9 said:
must be missing something, because I don't really see another way to solve it.

Thanks,
Matt

## 1. What is a critical point of a multivariable function?

A critical point of a multivariable function is a point where the partial derivatives of the function are equal to zero or do not exist. This means that the tangent plane to the surface of the function is either horizontal or undefined at that point.

## 2. How do you find the critical points of a multivariable function?

To find the critical points of a multivariable function, you need to take the partial derivatives of the function with respect to each variable, set them equal to zero, and then solve the resulting system of equations. The solutions to this system will be the critical points of the function.

## 3. Why are critical points important in multivariable functions?

Critical points are important in multivariable functions because they can give us information about the behavior of the function. They can indicate where the function has a maximum, minimum, or saddle point. They can also help us determine the shape of the function's graph.

## 4. Can a multivariable function have more than one critical point?

Yes, a multivariable function can have more than one critical point. In fact, it is common for multivariable functions to have multiple critical points. These points may represent local maxima, minima, or saddle points, depending on the behavior of the function in their respective neighborhoods.

## 5. How do you determine the nature of a critical point in a multivariable function?

To determine the nature of a critical point in a multivariable function, you can use the second derivative test. This involves taking the second derivatives of the function and evaluating them at the critical point. If the second derivative is positive, the critical point is a local minimum. If the second derivative is negative, the critical point is a local maximum. If the second derivative is zero, the test is inconclusive and you may need to use other methods to determine the nature of the point.