# $f(x) = \sqrt{9-x^2}$. Why can't the range be $\pm 3$?

In here, I understand the logic behind domain is $$[-3,3]$$ and the range is $$[0,3]$$ but if we try finding the range in algebraic way like the following,

$$y = \sqrt{9-x^2}$$

$$y^2 = 9-x^2$$

So this means $$y = \pm \sqrt{9-x^2}$$.

Could someone please explain why this reasoning is wrong?

Thanks.

• $a=b \implies a^2=b^2 \not\implies a=-b$ – Alfredo May 22 '19 at 10:51
• Shouldn't the title read: Why can't the range be $[-3,3]$? – Ramanujan May 22 '19 at 11:06

The range of the square root is defined to be non-negative. It can never go below $$0$$. In this case, we start with setting $$y$$ to be the value of some square root. Thus by definition $$y$$ cannot be negative.

When you square both sides, you lose that piece of information. So if we only had $$y^2 = 9-x^2$$ to work with, $$y$$ could be either positive or negative. Thus when you take the square root on both sides again, you correctly get $$y = \pm\sqrt{9-x^2}$$ (which is shorthand for "$$y$$ is either equal to $$\sqrt{9-x^2}$$, or to $$-\sqrt{9-x^2}$$").

What you have done here is to prove that if $$y$$ is equal to $$\sqrt{9-x^2}$$, then $$y$$ is equal either to $$\sqrt{9-x^2}$$ or to $$-\sqrt{9-x^2}$$. This should not be surprising.

For the function $$f(x) = \sqrt {9-x^2}$$ the domain is the set of points where the function is well defined and that is the set of $$x$$ values where $$9-x^2 \ge 0$$ That gives us $$|x|\le 3$$ or $$[-3,3]$$

The range is the set of all $$f(x)$$ as $$x$$ takes all the values in the domain and it is $$[0,3]$$

You have mentioned something about squaring the function and taking square root and considering two cases of plus or minus the square root.

That is not valid because you are altering your function by this process.

• There was an error in the title. The function was intended to be $f(x) = \sqrt{9 - x^2}$. – N. F. Taussig May 22 '19 at 11:03
• Thanks, I have edited my answer accordingly. – Mohammad Riazi-Kermani May 22 '19 at 11:08

The function is defined for all $$x$$ such that $$9-x^2 \ge 0.$$ We have $$9-x^2 \ge 0 \iff |x| \le 3.$$