Solving f(x,y) and Sin(x) + cos(y) / 2x - 3y

[Ravic85]

Homework Statement

A: f(x,y) = e^(x+y) + cos(xy) / Ln(xy)

b: Sin(x) + cos(y) / 2x - 3y

Homework Equations

I'm not exactly sure if I'm doing this correctly, my book a little vague.

The Attempt at a Solution

for A I have when x = -1 and y = any real real number

for B I have when x = 3 and y = 2

 

[dynamicsolo]

Why restrict the sets so much? The functions will not be continuous anyplace where the functions are undefined (or where the limit is not defined, or where the limit does not equal the function value).

The numerators are always defined, so they do not present a problem. So the ratios are undefined if either the denominator functions are undefined, or where the denominator functions are zero.

 

[Ravic85]

So do you mean on say part A I'd have x = any negative number? I don't quite understand how I would lessen the restrictions on part B.

 

[dynamicsolo]

Remember that you want the sets of points (x,y), so not everything hinges on x alone.

part A: Where would ln(xy) be undefined? Where is it zero?

part B: 2x - 3y is always defined, but where is this difference equal to zero?

 

[Ravic85]

Where Ln would be undefined is where you get a negative number? when it's zero it's ln(1)? And with B: I thought I answered where the difference is equal to zero with x = 3 and y = 2

 

[I like Serena]

Yes ln is undefined for negative numbers and also for zero.
Furthermore the nominator is zero when it's ln(1).

So any x and y for which xy<0 or xy=0 or xy=1 qualify for the function to be not continuous.

As it is you have given 1 example, but the question asks for all points (x,y) that qualify.

With B: For instance x=6 and y=4 also qualifies.

Edit: Oops. Fixed to y=4.

 

[dynamicsolo]

For part B, why not the entire line 3y = 2x --> y = (2/3) x ?

With B: For instance x=6 and y=2 also qualifies.

Do you mean x = 6 and y = 4?