1. ## Multivariable function fun;

Similar to the parametric question I recently posted, I need to find a value of c such that a piecewise multivariate function is continuous over an interval, not too sure where to start, because c is not only in the functon, but in the range.

Find all values of c, such that
$f(x,y) =\left\{\begin{array}{cc}\frac{x}{y+c},&\mbox{ if }
x \in R, y \geq 1-c\\xy, & \mbox{ if } x \in R, y< 1-c\end{array}\right.$

is continuous on $R^2$.
Not too sure where to start here, any pointers in the right direction?

2. I approach these questions by assuming that the two equations are equal at the point where the graph shifts from one to the other. This follows from the fact that a continuous function must have for each f(x,y) a limit that equals this value as well from either side.

So set them equal, $\frac{x}{y+c}=xy$

The x's cancel which makes sense because the restriction is dependent on y, not x.

Now this equality isn't true for all (x,y), it is only at the point around y=1-c. So if you plug in 1-c for y and solve for c, you should get the desired value.

3. So in order for them to be continuous they must equal each other, correct?

My question is, what happens to the inequalities? Can we really make them = to y? Why can we do that, does it have something to do with making the two equations equal to each other?

4. Originally Posted by Jameson
I approach these questions by assuming that the two equations are equal at the point where the graph shifts from one to the other. This follows from the fact that a continuous function must have for each f(x,y) a limit that equals this value as well from either side.

So set them equal, $\frac{x}{y+c}=xy$

The x's cancel which makes sense because the restriction is dependent on y, not x.

Now this equality isn't true for all (x,y), it is only at the point around y=1-c. So if you plug in 1-c for y and solve for c, you should get the desired value.
I like that rationale, I'm supposing that plugging in the 1-c for y works because we would be subbing it in on either side regardless to take the left and right hand limit, but I'm getting a peculiar answer by doing that.

$\frac{x}{y+c}=xy$

$\frac{1}{y+c}=y$

$1= y^2 +yc$

$1=(1-c)^2 +c(1-c)$

$1=1-c$

$c=0$

Which doesn't make sense because when we sub that back into the original statement, $\frac{x}{y+c}=xy$, we get $\frac{x}{y}=xy$?

I must have made a mistake somewhere.

5. Originally Posted by Orbent
So in order for them to be continuous they must equal each other, correct?
In order for them to be continuous, the limits must be the same. Since the two functions that are used are continuous functions, the limits are the same as value of the functions.

[/quote]My question is, what happens to the inequalities? Can we really make them = to y? Why can we do that, does it have something to do with making the two equations equal to each other? [/QUOTE]
Again, it is a matter of limits. The limit as some point (x, y) approaches (x, 1- c) (the boundary between $y\ge 1- c]$ and $y< 1- c$) depends on what happens on either side of y= 1- c.

6. Originally Posted by Kasper
I like that rationale, I'm supposing that plugging in the 1-c for y works because we would be subbing it in on either side regardless to take the left and right hand limit, but I'm getting a peculiar answer by doing that.

$\frac{x}{y+c}=xy$

$\frac{1}{y+c}=y$

$1= y^2 +yc$

$1=(1-c)^2 +c(1-c)$

[tex]1=1-c[/math

$c=0$

Which doesn't make sense because when we sub that back into the original statement, $\frac{x}{y+c}=xy$, we get $\frac{x}{y}=xy$?

I must have made a mistake somewhere.
Your "original statement", $\frac{x}{y+c}= xy$, is only true on the line y= 1- c which is y= 1 since c= 0. Yes, that becomes $\frac{x}{1}= x(1)$!

$
f(x,y) =\left\{\begin{array}{cc}\frac{x}{y},&\mbox{ if }
x \in R, y \geq 1\\xy, & \mbox{ if } x \in R, y< 1\end{array}\right.
$

which is continuous for all (x, y).

7. Originally Posted by HallsofIvy
Your "original statement", $\frac{x}{y+c}= xy$, is only true on the line y= 1- c which is y= 1 since c= 0. Yes, that becomes $\frac{x}{1}= x(1)$!

$