Question on finding the range of a function involving squareroot The question is as under :


  Find the range of the real valued function : $f(x) = \dfrac{1}{\sqrt{16-x^2}}$


The answer in the book is :  $(-\infty,1/4] \cup [1/4, \infty)$.

But according to me the answer should be only $[1/4, \infty)$ because $f(x)$ cannot take negative values because if it takes negative values,
there will be two values of $f(x)$ corresponding to one value of $x$ and then $f(x)$ will not remain a function.

Then why is the answer in the book different?

Please give me the explanation. Thank you.
You are correct. 

$\sqrt{16-x^2}$ only takes on positive values from $(-4,4)$, which range from $0$ to $4$.
Well, the zeros of $16-x^2$ are $\pm 4$ and the expression $16-x^2$ is non-negative if  $-4\leq x\leq 4$. Now stick it together to find the real-valued solutions of $1/\sqrt{16-x^2}$.