# Range of square root of rational function

How can I determine the range of the function $$f(x)=\sqrt{\frac{-x^2+2x+3}{18x-3x^3}}$$ without using limits or derivatives? I have factorised numerator and denominator, but nothing simplifies. I tried solving $$y=f(x)$$ for $$x$$, because the domain of the inverse function is the range of the initial function, but I am somehow stuck. My guess is that the range is all non-negative real numbers. How can I be sure that we reach all of them? If we have it for the fraction, we will also have it for the whole root. Can someone help me here? Thanks!

• Can you show that, for each $M>0$, there exists an $\varepsilon \in (0, \frac{\sqrt{6}}{2})$, such that $f(\sqrt{6} - \varepsilon) > M$? This is cheating a bit, because it's essentailly finding the limit at $\sqrt{6} - 0$, but that's the only way to deal with infinity. Oct 30 '20 at 21:44

First the denominator has degree 3 while the numerator has degree 2. For very large x, this is close to 0 so the minimum is 0. The denominator is 0 for x= 0, $$\sqrt{3}$$, and $$-\sqrt{3}$$ while the numerator is not 0 for those x values so the function value can be arbritrarily large. (I hope I haven't come too close to "limits" for you?)

Let $$M \in (0, +\infty)$$. $$M=0$$ is obviously hit, for example by $$x = 3$$.

We want to solve the equation $$\sqrt{\frac{x^2 - 2x - 3}{3x^3 - 18x}} = M.$$Squaring and then transforming the expression gives us $$3M^2 x^3 - x^2 + (2-18M^2)x + 3 = 0.$$For $$M > 0$$ this is a cubic equation, so it has at least one real root. Let $$x_0$$ be such a root. Obviously $$x_0 \notin \{0, \sqrt{6}, -\sqrt{6}\}$$, so $$\frac{x_0^2 - 2x_0 - 3}{3x_0^3 - 18x_0}$$ is a well-defined number equal to $$M^2$$, so it's positive. Therefore, its square root is defined and equal to $$M$$.

• I showed that for every $M>0$, the root $x_0$ satisfies $f(x_0) = M$. Oct 30 '20 at 22:29

$$\frac{(x+1)(3-x)}{3x(\sqrt6-x)(\sqrt6+x)}$$ must be non-negative. Notice that for $$x$$ in

$$(-\sqrt6,-1]$$ this fraction takes all possible non-negative values, as the signs are

$$\frac{-+}{-++}$$ and the denominator tends to $$0$$ on the left, and the numerator to $$0$$ on the right.

Hence, $$\mathbb R^+.$$

Find an interval in which the radicand, and thus $$f$$ itself is continuous (for example, for all $$x$$ such that $$-\sqrt 6).

In this interval, show that $$f$$ vanishes. This happens for example for $$x=-1.$$ Then the result follows immediately.