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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!

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  • $\begingroup$ 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. $\endgroup$ Oct 30 '20 at 21:44
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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?)

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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 $M$.

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  • $\begingroup$ I showed that for every $M>0$, the root $x_0$ satisfies $f(x_0) = M$. $\endgroup$ Oct 30 '20 at 22:29
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$$\frac{(x+1)(3-x)}{3x(\sqrt6-x)(\sqrt6+x)}$$ 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, $$\mathbb R^+.$$

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Find an interval in which the radicand, and thus $f$ itself is continuous (for example, for all $x$ such that $-\sqrt 6<x<0$).

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

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