The function is:
$f(x,y) = \frac{2}{\sqrt{3-x}} + \frac{1}{\sqrt{4-y}}$
I have found the domain and the Range intuitively.
But how would I formally prove that my assumption of the Range and Domain is true?
The function is:
$f(x,y) = \frac{2}{\sqrt{3-x}} + \frac{1}{\sqrt{4-y}}$
I have found the domain and the Range intuitively.
But how would I formally prove that my assumption of the Range and Domain is true?
On
You can input Your function in WolframAlpha
From alternate form the domain and range is quite clear.
On
I suppose the function is defined on $ \mathbb{R}^2 $; thus the domain has to contain elements of $ \mathbb{R}^2 $ such that the function gives a result in $ \mathbb{R} $. For the range, you can observe that for every element in the domain of $f$ you have $f(x,y)>0$, then you notice that as $ x \rightarrow 3 $, $ y \rightarrow 4 $, $ f(x,y) $ takes arbitrary large values. Since the function is continuous on its domain, the range is thus $ (0,+\infty)$.
Informally, to find the domain, we simply examine sub-parts of the expression, looking for potentially illegal operations. When we see a sub-part like $\sqrt{4-y}$, for example, we know that $4-y$ better be positive. We find all such restrictions and then form their conjunction.
As far as agnis's WolframAlpha suggestion goes, it's hard to imagine a more "formal" computation than what the computer might do. Your function
can be represented as a graph theoretical tree:
We now traverse the tree. Some nodes lead to restrictions; others do not. For example, a
Plusnode is no difficulty, since we can always add any two real numbers. We just record the visit and move on. APowernode can lead to a problem. Specifically,Power[expression,exponent]leads to the restrictionexpression!=0, wheneverexponent<0. Furthermore,expressionmust be non-negative ifexponentis a rational number. Thus, for example, thePowernode on the right leads to the inequality $4-y>0$. Finally, we take the intersection of all these parts.