What is the type of discontinuity at the origin for the function 2x^y/x^4+y^2?

In summary, a multivariable function is a mathematical function that takes in multiple variables as inputs and produces a single output. It is different from a single variable function in that it can take in multiple input variables, allowing for a more complex relationship between the input and output. To graph a multivariable function, a coordinate system with multiple dimensions, such as a 3-dimensional graph, is needed. Studying multivariable functions is important in various fields and can help in solving complex problems. To find the critical points of a multivariable function, the partial derivatives of the function with respect to each variable must be set equal to zero.
  • #1
kazthehack
5
0
1.)

Homework Statement


Sketch the domain of f(x,y,z) = ln(25-x^2-y^2-z^2) and determine it's range.

Homework Equations


N/a

The Attempt at a Solution


Im sure the domain is x^2+y^2+z^2 < 25 or (-infinity, 25)
Then the range is (0,+infinity.)
my problem is how would i sketch it.. i was thinking of a sphere with dotted outline at 5,5,5

2.

Homework Statement


Specify the type of discontinuity at the origin of the following functions.

2x^y/x^4+y^2

Homework Equations


N/a

The Attempt at a Solution


i am not sure how would i find the limit of f(x,y) as (x,y)->(0,0) so i can prove that it is either essential or removable discontinuity..

any help would be great thanks! :D
 
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  • #2
kazthehack said:
1.)

Homework Statement


Sketch the domain of f(x,y,z) = ln(25-x^2-y^2-z^2) and determine it's range.

Homework Equations


N/a

The Attempt at a Solution


Im sure the domain is x^2+y^2+z^2 < 25 or (-infinity, 25)
The domain is {(x, y, z) | x^2+y^2+z^2 < 25 }. This domain is a subset of three-dimensional space, not an interval as you show it.
kazthehack said:
Then the range is (0,+infinity.)
No. Think about the largest and smallest values that ln(25-x^2-y^2-z^2) can attain. Sure, it will always be > 0, but there is a finite value that it is always less than.
kazthehack said:
my problem is how would i sketch it.. i was thinking of a sphere with dotted outline at 5,5,5
You're on the right track with the sphere, as the domain is the interior of some sphere, but the point (5, 5, 5) is outside the sphere. I don't understand what you're trying to say with "dotted outline at 5,5,5."
kazthehack said:
2.

Homework Statement


Specify the type of discontinuity at the origin of the following functions.

2x^y/x^4+y^2
Please write this more carefully so I can clearly see what's in the denominator. Most people on this forum would interpret this as 2(xy/x4) + y2, but I suspect that isn't what you mean.
kazthehack said:

Homework Equations


N/a

The Attempt at a Solution


i am not sure how would i find the limit of f(x,y) as (x,y)->(0,0) so i can prove that it is either essential or removable discontinuity..

any help would be great thanks! :D
 
  • #3
So as for the first question the graph would be a sphere that is shaded inside? and then the outline of the sphere would be dotted to indicate that it wouldn't reach the radius of 5.

As for the 2nd one ...
the equation was 2x^2y/(x^4+y^2)
 
  • #4
kazthehack said:
So as for the first question the graph would be a sphere that is shaded inside? and then the outline of the sphere would be dotted to indicate that it wouldn't reach the radius of 5.
Do you mean the graph of the domain of the function or the graph of the function? The domain of your function is the interior of a sphere centered at (0, 0, 0) and of radius 5.

I haven't discussed the graph of the function, other than to say something about its domain and range.
kazthehack said:
As for the 2nd one ...
the equation was 2x^2y/(x^4+y^2)
Well, that's what I though you meant, but I wanted to get you to write it so that it would be generally understandable.

For the limit as (x, y) --> (0, 0) of this function to exist, it must exist independent of the path taken. Try approaching (0, 0) along different lines (y = kx) and different curves, and see if they come out the same. For different curves, you might try y = x2, y = x3, and so on.
 
  • #5
Mark44 said:
No. Think about the largest and smallest values that ln(25-x^2-y^2-z^2) can attain. Sure, it will always be > 0, but there is a finite value that it is always less than.
Can you clarify this?
 
  • #6
kazthehack said:
Can you clarify this?

25-x^2-y^2-z^2=25-(x^2+y^2+z^2). What's the largest value that can ever be?
 

What is a multivariable function?

A multivariable function is a mathematical function that takes in multiple variables as inputs and produces a single output. Each variable in the function represents a different factor that can affect the output.

What is the difference between a multivariable function and a single variable function?

A single variable function only takes one input variable, whereas a multivariable function takes in multiple input variables. This allows for a more complex relationship between the input and output.

How do you graph a multivariable function?

In order to graph a multivariable function, you will need to use a coordinate system with multiple dimensions, such as a 3-dimensional graph. Each axis represents a different input variable, and the output is represented by the height or depth of the graph at a given point.

What is the importance of studying multivariable functions?

Studying multivariable functions is important in many fields, such as physics, economics, and engineering. It allows for a more accurate representation of real-world phenomena and can help in making predictions and solving complex problems.

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

To find the critical points of a multivariable function, you will need to take the partial derivatives of the function with respect to each variable and set them equal to zero. The solutions to these equations will give you the critical points, which can then be used to determine the maximum and minimum values of the function.

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