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The question is: find the domain and range of $f(x) = \sqrt{{(1-x)}/x}$.

For the domain, I took $x ≠ 0$ and $(1-x)/x ≥ 0$, which gave me $1 ≥ x$.

So the domain I found is $(-∞,0)∪(0,1]$. However, the proposed solution says that the domain is just $(0,1],$ without explaining, so I wonder why.

Also, any quick tip to find the range (without taking monotonicity, roots, limits to infinity etc.)? I tried inverting the function and the domain of the inverse is $\Bbb R$, which is clearly not the range of $f$.

Thanks in advance!

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  • $\begingroup$ What happens to the inequality sign when you multiply by negative numbers? This is the reason for the domain. $\endgroup$
    – Metric
    Jan 9, 2019 at 0:07
  • $\begingroup$ I know the inequality changes when we multiply it by a negative number, however my computation was this: (1-x)/x ≥ 0 implies 1/x -1 ≥ 0 implies 1/x ≥ 1 implies 1 ≥ x, so I don't think I multiplied by a negative number anywhere. Also, the module only includes real functions of 1 variable, so "the natural domain" kind of means "Find all real x such that the output is real and defined" I guess. Feel free to point any further mistakes. $\endgroup$
    – user600210
    Jan 9, 2019 at 0:14
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    $\begingroup$ This being an algebra-precalculus question, I know that the formality behind "domain" is watered down, so I know what you meant in your question. Your process works only for when $x > 0$. When $x < 0$, the inequality sign would flip. $\endgroup$
    – Metric
    Jan 9, 2019 at 0:17
  • $\begingroup$ After it "flips", what resultant inequality would you get? $\endgroup$
    – Metric
    Jan 9, 2019 at 0:18
  • $\begingroup$ Ohh I just saw it... my mind is not functioning properly today I guess. Any helpful tips on how to find the range? $\endgroup$
    – user600210
    Jan 9, 2019 at 0:32

2 Answers 2

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$f(x) = \sqrt{\frac {1-x}x}$ which is undefined if $x =0$ and if $\frac {1-x}x = \frac 1x - 1 < 0$. $x\ne 0$ and $\frac 1x -1 < 0 \implies \frac 1x \ge 1$.

If $x > 0$ then $\frac 1x \ge 1 \implies x \le 1$ and if $x < 0$ implies $x \ge 1$ which is a contradiction so $0 < x \le 1$ and domain is $(0,1]$.

[Alternatively $\frac 1x \ge 1 > 0$ so $\frac {\frac 1x}{\frac 1x} \ge \frac 1{\frac 1x}>\frac 0{\frac 1x} \implies 1 \ge x > 0$.]

As for range.

Now that we know $0 < x \le 1$ then $\frac 1x \ge 1$ and $\frac 1x - 1 \ge 0$ and $\frac {1-x}x = \frac 1x -1 \ge 0$ and $\sqrt{\frac {1-x}x} \ge 0$ with no restrictions.

That implies the range is $[0, \infty)$.

But just to make certain we should test if $\sqrt{\frac {1-x}x} = m$ is solvable for all $m \ge 0$.

$\sqrt{\frac {1-x}x} = m \ge 0 \implies$

$\frac{1-x}x =\frac 1x- 1= m^2 \implies$

$\frac 1x = m^2 + 1$. So $m^2 + 1 \ge 1$ we can conclude

$x = \frac 1{m^2 + 1}$ and that exists for all possible $m\ge 0$

So range is all $[0, \infty)$

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So the domain I found is $(-∞,0)∪(0,1]$. However, the proposed solution says that the domain is just $(0,1],$ without explaining, so I wonder why.

Suppose you take $f(x)$ for $x < 0$.

Then you know $1 - x > 0$, and $x < 0$.

Thus, since dividing a positive by a negative is a negative,

$$\frac{1-x}{x} < 0$$

Since you're contending with a function of real values here, you can't have the radicand be negative. That's why the domain of $f$ is only $(0,1]$.


Also, any quick tip to find the range (without taking monotonicity, roots, limits to infinity etc.)? I tried inverting the function and the domain of the inverse is $\Bbb R$, which is clearly not the range of $f$.

Whenever easily doable, as noted, taking the inverse is the way to go. However, I believe you made a slight oversight in investigating said inverse. I'll start from the beginning for completion's sake.

Let

$$y = \sqrt{ \frac{1-x}{x} } = \sqrt{ \frac{1}{x} - 1}$$

Solving for $x$ (i.e. getting the inverse) yields

$$x = \frac{1}{y^2 + 1}$$

Now, in this context, "range" means the set of values $f$ maps to on its domain. In other words, since the domain of $f$ is $(0,1]$, we're looking at the inverse for which $y$ values map to that interval.

You can easily confirm this by looking at either the graph of $f$ or $f^{-1}$ depending on the interpretation you want to fiddle with. But, analytically, we note:

  • $0 < x \leq 1$ gives us our inputs $x$.
  • Can $f^{-1}$ be greater than $1$? (Hint: notice: $y^2 \geq 0$.)
  • Can $f^{-1}$ be less than $0$? (Hint: notice from the previous that the numerator and denominator are always positive.)

Depending on how formal/informal your argument could be, utilizing the previous and the fact that we're looking at the image of $(0,1]$ under $f$ might be sufficient. Again, graphs are also useful for all this. (They're not a substitute for a proof, but they can be very useful in seeing what to do.)

Of course, we conclude that the above seems to imply the image of $f$ is $\Bbb R$ since we never really restricted the values of $y$. All the above really seem to show is that any $y$ can give us an $x$ in our domain.

(This might be touching on your error: it's always useful to reconsider, after analyzing the inverse, to see if the values you found up to now can be mapped to.)

However, if $y$ is in the image of $f$, that means there exists some $x$ so that $f(x) = y$, right? Then suppose $y<0$ - it should be immediately obvious why the negative real numbers are not in the image of $f$, namely because we take the principal square root, i.e. $f$ is always positive or zero.

So with these observations in mind, it becomes clear then that the range of $f$ is not all real numbers - just the nonnegative ones.

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    $\begingroup$ Thanks for that... I usually mindlessly find the domain of the inverse of f, and that's what I did here as well, obviously omitting crucial stuff. Thanks! (Note: the calculation of the monotonicity, the right-handed limit of f to 0 and f(1), while a bit more tedious, would have given me the correct range, so I guess i shot my own foot there) $\endgroup$
    – user600210
    Jan 9, 2019 at 0:53
  • $\begingroup$ lol understandable, it took me a second to figure it out too. And yeah, those arguments are valid too and probably way more formal than my lame heuristic appeal to intuition. :p $\endgroup$ Jan 9, 2019 at 0:55

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