1 Answer | Add Yours
This question seems a bit interesting to me in that its apparent answer is actually pretty uninteresting!
The question we must ask ourselves: When might this function be non-continuous? Well, we need to figure out what the domain of the function is, and whether any of the possible values of (x,y) go outside that domain. Then, we need to show where there are "jumps" or other sorts of discontinuities that would make it noncontinuous.
Let's answer the first question: What values of (x,y) lie outside the domain of the function?
Let's look at the function, itself:
`ln(4x^2+9y^2 + 36)`
The only type of function here in which the domain would be limited is the logarithm. Its domain is limited to the positive real numbers!
So, we just need to see where the input of the logarithm will go to 0 or negative:
`4x^2 + 9y^2 + 36 <= 0`
We can simplify (recall: `i = sqrt(-1)` )``:
`x^2/(3i)^2 + y^2/(2i)^2 >= 1`
Well, it looks like we have an ellipse in the complex plane! However, this is only a worry for us if `(x,y) in CC^2`. Now, if what I just said doesn't make sense, don't worry, because unless you're dealing with math-major-ish math, this is something you likely don't need to worry about.
So, what did we also show with the above ridiculous equation? That if `(x,y)inRR^2`(i.e. if x and y are real numbers) then the domain for our function `f(x,y)` is the entire x and y axes.
Therefore, we aren't going to find discontinuities by looking for where x and y may lie outside the domain of the function.
Now, we check to see if there are any jumps. Recall, a function's domain may include all possible (x,y), but there still may be points at which there may be discontinuities. For example, think of a step function. The domain is the entire x-axis, but the function is not continuous at the step! Therefore, we need to rule out these sorts of discontinuities!
Well, in our case, this is simple. Because there is no piecewise nature of the function above, and because there is no (x,y) that brings it out of its domain, it will be continuous at all points. Another way to do this is through taking partial derivatives:
`(delf)/(delx) = (8x)/(4x^2+9y^2+36)`
`(delf)/(dely) = (18y)/(4x^2+9y^2+36)`
If you look at these partial derivatives, you can see that, just like for the logarithm above, the are defined for all (x,y). Nowhere can their denominator become 0 (unless `(x,y)inCC^2` but again, I don't think we need to worry about this!), indicating that the derivatives for f exist everywhere. Recall that in order for a function to be differentiable at each point, it also needs to be continuous at each point. Therefore, f is continuous at all (x,y).
So, there are a couple ways of showing this. I hope that helped out!
Again, if the complex numbers threw you off, don't worry about them! That is there purely for enrichment and almost certainly not what the problem is after.
We’ve answered 319,812 questions. We can answer yours, too.Ask a question