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Consider the surface given by the function $f(x, y) = \frac{4}{\sqrt{x^2 + y^2 -9}}$. What is the range of $f$? Determine any $y$-intercept(s) of $f$.

For range, I realize that $\sqrt{x^2 + y^2 -9}$ can be any number, so long as $x^2 + y^2 -9$ is positive or equal to zero. However, when looking at $f$, we know that the denominator cannot equal zero, which means that $\sqrt{x^2 + y^2 -9}$ can be any number greater than 0. Would this mean that the range of $f$ is $(0, \infty)$?

For the $y$-intercept, I know that both $x$ and $z$ need to equal zero. Would this mean that there are no $y$-intercepts because 0 is not in the range of $f$?

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There's one step missing in your computation of the range. Yes, numbers of the form $\sqrt{x^2+y^2-9}$ can take any value in $(0,+\infty)$. You should have added that it follows from this that then the same thing occurs with $\frac4{\sqrt{x^2+y^2-9}}$.

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  • $\begingroup$ Thanks, I’ll add that in. Is my interpretation of the y-intercept correct? $\endgroup$ – 631 Jun 17 '18 at 18:43
  • $\begingroup$ @631 Yes, it is. $\endgroup$ – José Carlos Santos Jun 17 '18 at 18:45

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