MHB Mr.Ask's question at Yahoo Answers (Two variables, domain and range)

Click For Summary
The function f(x,y) = 1/√(16 - x² - y²) is defined only when 16 - x² - y² > 0, which corresponds to the domain being the open disk D((0,0),4). Therefore, the domain of f is the set of all points (x,y) such that x² + y² < 16. The range of the function varies from 1/4 to infinity, with 1/4 included and infinity excluded. Thus, the range is [1/4, +∞). This analysis provides a clear understanding of the function's domain and range.
Fernando Revilla
Gold Member
MHB
Messages
631
Reaction score
0
Mathematics news on Phys.org
Hello MrAsk,

There exists $f(x,y)=\dfrac{1}{\sqrt{16-x^2-y^2}}$ if and only if $16-x^2-y^2>0$ or equivalently if and only if $x^2+y^2<4^2$ (open disk $D((0,0),4)$). That is, $$\boxed{\:\mbox{Dom } (f)=D((0,0),4)\:}$$ On the other hand, for every $(x,y)$ such that $x^2+y^2=r^2$ with $0\leq r<4$ we have $f(x,y)=\dfrac{1}{\sqrt{16-r^2}}$ so, $f(x,y)$ varies from $1/4$ (included) to $+\infty$ (excluded). That is, $$\boxed{\:\mbox{Range } (f)=\left[1/4,+\infty\right)\:}$$
 
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
1K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K