# 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.

• That is not a problem. That will imply the function is not surjective – user665856 May 1 at 7:17
• Rest assured, $f$ cannot take negative values and the book is wrong . – Kavi Rama Murthy May 1 at 7:36

You are correct.

$$\sqrt{16-x^2}$$ only takes on positive values from $$(-4,4)$$, which range from $$0$$ to $$4$$.

• @Allawonder i said positive values. -4, and 4 are zeroes – Saketh Malyala May 1 at 8:13
• Thanks , but I have done many other questions in that book like finding the range of √(16−x^2). Then the answer in the book is [0,4]. Why didn't they consider negative values in this case. – Ashok Sharma May 1 at 14:39
• @SakethMalyala But $0$ is neither positive nor negative, except you're French of course. – Allawonder May 1 at 16:27

You say that

$$f(x)$$ 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.

Since $$f(x)$$ is defined by the expression $$\frac{1}{\sqrt{16-x^2}},$$ I agree with you that it cannot possibly be negative, or vanish for that matter. Also, you're right that this would make $$f(x)$$ double-valued at some points. Another way to see this is to understand that the symbol $$\sqrt{}$$ means the nonnegative square root of, so that $$\sqrt x$$ is uniquely defined whenever $$x$$ is not negative. Thus, I also agree that the range of the function is $$[1/4,\infty).$$

To answer your question, the answer in your book may be different perhaps because at that level they simply want you to be able to manipulate with symbols and know what processes to follow -- which is a bad pedagogical method, in my opinion. Most books at secondary level are also cavalier about precision and pay little attention to such matters, which are of more importance than mere calculation. When one does the calculation, it's true that what drops out is the answer given by your author, namely $$(-\infty,-1/4]\cup[1/4,\infty),$$ but we also have to remember that $$f(x)$$ by definition cannot be negative.

Finally, in case of next time, don't be shy to think that an author might be wrong. An author is also human, so it's possible. The only caveat is that if it's a well-established book gone through more than a couple of editions with a reputable publisher, then the possibility is minimal indeed, so that one should be careful. However, it always remains a possibility. No book is perfectly error-free.

• Thanks , but I have done many other questions in that book like finding the range of √(16−x^2). Then the answer in the book is [0,4]. Why didn't they consider negative values in this case. – Ashok Sharma May 1 at 14:34
• @AshokSharma Well, in that case the only rational possibility is oversight. As I said in my post, even the most carefully prepared book isn't completely free of all kinds of error. By the way, check other examples of this kind in that book; their most consistent solutions will ascertain whether it's indeed an oversight in this isolated case, or is typical of them. – Allawonder May 1 at 16:33
• Ok . If I am doing any such type of question involving square root then should I take ONLY the non negative values of y in the range – Ashok Sharma May 1 at 17:11
• Well, by definition, $\sqrt x\ge 0$ provided $x\ge 0,$ too. Thus, $\sqrt x$ cannot be negative -- ever! If you want negative square roots, you write $-\sqrt x,$ for example. – Allawonder May 1 at 17:54

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 $$1/\sqrt{16-x^2}$$.