I have the following multivariable function: f(x,y) = xy^2, and

Arely Briggs

Arely Briggs

Answered question


I have the following multivariable function: f(x,y)=xy2, and I must prove it is continuous at (1,1). I have come to: |((x1)2+(y1)2)321|((x1)2+(y1)2)32+1 (Triangle inequality) <δ3+1=ϵ.
What can I do with this 1? Any tips are welcome and thank you in advance!

Answer & Explanation

Addisyn Thompson

Addisyn Thompson

Beginner2022-01-26Added 16 answers

One can pose this problem by considering the graph of the function as follow.
From the definition of continuity the ε-neighbourhood of the image f(1,1)=1 determines a couple of planes z=1+ε and z=1ε
The intersections of the planes 1±ε with the graph of f determine two hyperbolas
with xy2=1±ε.
Also a little circle of radius δ, which is the border of a δ-neighbourhood centered at the preimage (1,1), when is transformed gives a closed curve C viewed in the graph of f.
Then there is a point in the curve C which is the farest away from p=(1,1,1) , which will be determined by considering that for the curve C can be parametrized as C(y)=(1+εy2,y,1+ε) and the distance function

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Multivariable calculus

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?