Continuity of g(x,y) = (xy)^1/3

  • Thread starter pergradus
  • Start date
  • Tags
    Continuity
In summary, The function g(x,y) = (xy)1/3 is continuous at the point (0,0), as shown by the limit as (x,y)->(0,0) and the function being equal to 0 at (0,0). However, some graphing programs may not accurately display the function due to the cube root. The ultimate goal is to determine the differentiability of the function at (0,0).
  • #1
pergradus
138
1

Homework Statement



Show if the function g(x,y) = (xy)1/3 is continuous at the point (0,0)

Homework Equations





The Attempt at a Solution



I'm a bit confused. When I take the limit as (x,y)->(0,0) I get that L = 0, and the function is equal to 0 at (0,0), but when I plot the function in maple, it seems as if there is a discontinuity along the line x =0 and y = 0. I am ultimately trying to show if the function is differentiable at (0,0). Any help?
 
Physics news on Phys.org
  • #2
Well, your calculations are correct. The function is continuous.

Many function graphers cannot handle the cube root well, which is likely the reason that it doesn't show up nice in maple.
 

1. What is the definition of continuity?

The definition of continuity for a function at a given point is that the limit of the function as x and y approach that point must equal the value of the function at that point.

2. Is g(x,y) = (xy)^1/3 continuous at (0,0)?

Yes, g(x,y) = (xy)^1/3 is continuous at (0,0) because the limit of the function as x and y approach (0,0) is equal to 0, which is also the value of the function at (0,0).

3. What is the limit of g(x,y) = (xy)^1/3 as x and y approach (2,3)?

The limit of g(x,y) = (xy)^1/3 as x and y approach (2,3) is 2, because (2,3) is a continuous point on the function and the value of the function at that point is 2.

4. Can g(x,y) = (xy)^1/3 be written as a piecewise function?

Yes, g(x,y) = (xy)^1/3 can be written as a piecewise function where it is defined by (xy)^1/3 when x and y are both positive, -(-xy)^1/3 when x and y are both negative, and 0 when x and y have opposite signs. This is to ensure continuity at (0,0).

5. What is the partial derivative of g(x,y) = (xy)^1/3 with respect to x?

The partial derivative of g(x,y) = (xy)^1/3 with respect to x is 1/3(yx^2)^-2/3 * y = y/(3x^2(yx^2)^1/3).

Similar threads

Verify Green's Theorem in the given problem
    • Calculus and Beyond Homework Help
Replies
10
Views
445
Continuity of ##g(x,y)## and its partials
    • Calculus and Beyond Homework Help
Replies
23
Views
2K
Show that the given function is continuous
    • Calculus and Beyond Homework Help
Replies
27
Views
740
Prove if ##x<0## and ##y<z## then ##xy>xz## (Rudin)
    • Calculus and Beyond Homework Help
Replies
2
Views
596
Prove that f is a homeomorphism iff g is continuous, fg=1 and gf=1
    • Calculus and Beyond Homework Help
Replies
5
Views
1K
Find f(x,y) given partial derivative and initial condition
    • Calculus and Beyond Homework Help
Replies
6
Views
549
Is the given function continuous at x= 0? f(x) = {sin(1/x) if x≠0, 1
    • Calculus and Beyond Homework Help
Replies
4
Views
2K
Find the local maxima and minima for##f(x,y) = x^3-xy-x+xy^3-y^4##
    • Calculus and Beyond Homework Help
Replies
5
Views
548
Differentiability in higher dimensions
    • Calculus and Beyond Homework Help
Replies
11
Views
2K
A few questions to make sure I understand Fourier series
    • Calculus and Beyond Homework Help
Replies
3
Views
287

Hot Threads

Recent Insights

Back
Top