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Let $f(x)=$ $\dfrac{1}{\sqrt{x-5}}$ is a given function. We have to find its range.
I have tried two approaches:-

  1. $\sqrt{x-5}>0$
    ⇒ $\dfrac{1}{\sqrt{x-5}}>0$
    ⇒ $y>0$
    ⇒ Range = $(0,∞)$
  2. $f(x)=y=$ $\dfrac{1}{\sqrt{x-5}}$
    ⇒ $y^2$=$\dfrac{1}{{x-5}}$
    ⇒ $x=5+$ $\dfrac{1}{{y^2}}$
    ⇒ Range = R – {$0$}
    The ranges found are not the same.

My view is that, this happened because in the 2nd approach, when I squared both sides, the function is changed. And what I had calculated is the range of $y^2=\dfrac{1}{{x-5}}$ instead of $y=$ $\dfrac{1}{\sqrt{x-5}}$.

Is this statement correct?

amWhy
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Shekhar Malhotra
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  • In your second method, you need to note both $x\gt 5$, and that $y>0$. So $\operatorname{Range} = (0, \infty)$ because with $x\gt 5$, $x-5 \gt 0$, which requires $\frac 1{y^2} \gt 0$ which implies $y\gt 0$. – amWhy Mar 31 '20 at 15:44
  • @amWhy: I am not interested in the range. My statement is that, in the 2nd approach, what I have calculated, is actually the range of $y^2$=$\dfrac{1}{{x-5}}$ instead of $y=$ $\dfrac{1}{\sqrt{x-5}}$. Is this statement correct? – Shekhar Malhotra Mar 31 '20 at 16:08
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    Yes. correct... In your second approach, you did not calculate the range of $y=\frac{1}{\sqrt{x-5}}$. – amWhy Mar 31 '20 at 16:13
  • Again, yes, the answer to your most recent edit is yes, your conclusion is correct. – amWhy Mar 31 '20 at 16:39

3 Answers3

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The range is $(0,\infty )$. Moreover, your argument is not correct. For example, $|\arctan(x)|\geq 0$, but it's range is $[0,\frac{\pi}{2}]$.

The argument is if $f(x)=\frac{1}{\sqrt{x-5}}$, then $f(x)>0$ for all $x>5$ and $$\lim_{x\to \infty }f(x)=0 \quad \text{and}\quad \lim_{x\to 5}=+\infty .$$ Therefore, by Intermediate value theorem, $$\text{Im}(f)=(0,\infty ).$$

Surb
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1

You simply forgot, in your second approach, that $$A=\sqrt{\mathstrut B}\iff A^2=B \quad\textbf{and}\quad {A \ge 0} $$

amWhy
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Bernard
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    No, it's only because the symbol $\sqrt B$ denotes conventionally the nonnegative square root of a (nonnegative) number. The right side of the equivalence implies $B\ge 0$. – Bernard Mar 31 '20 at 15:29
  • Except in this case, $y = \frac 1{\sqrt{x -5}} \neq 0$. Clearly, $x\gt 5$, and $y\gt 0$. – amWhy Mar 31 '20 at 15:29
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    Of course, here, there's one more constraint. – Bernard Mar 31 '20 at 15:30
  • Indeed, $x\gt 5, y \gt 0$. So you oversimplified your general rule. It depends on B. $A= \sqrt B \iff A^62 = B$ and the definition of $B$, which determines what non-negative values $A$ can take on. – amWhy Mar 31 '20 at 15:31
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    This general rule supposes only $A$ and $B$ are (well-defined) real numbers. – Bernard Mar 31 '20 at 15:34
  • Well, in this case, $B$ is well defined for $x>5$, so me must have that A>0$. – amWhy Mar 31 '20 at 15:36
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    $B$ defined for $x>5$ ensures $A$ is defined, but $B$ has two square roots. Once again the symbol in use means the nonnegative square root (or the positive square root when one is sure it is not $0$), but this is purely a convention. – Bernard Mar 31 '20 at 15:40
  • @Bernard: Is this statement correct that 'in the 2nd approach when I squared both sides, the function is changed and what I have calculated, is actually the range of $y^2$=$\dfrac{1}{{x-5}}$ instead of $y=$ $\dfrac{1}{\sqrt{x-5}}$?' – Shekhar Malhotra Mar 31 '20 at 16:01
  • Well, it's hard to say, because $y^2$ cannot take any value since $y^2 can't be negative (if $y$ is real, of course), so it eliminates the negative numbers. In my opinion, you forgot the sign constraint – which, all in all, is more or less what I answered. – Bernard Mar 31 '20 at 16:18
  • $y^2$ term cannot be negative that is ok, but $y^2=\dfrac{1}{{x-5}}$ is a function which contains an even power of y, so it is symmetric about x-axis. Hence, it takes negative values too. – Shekhar Malhotra Mar 31 '20 at 18:07
  • $y$ can take negative values, not $y^2$. You seem to be confusing $y$ and $y^2$. – Bernard Mar 31 '20 at 18:09
  • No, I have asked that what I have calculated in the second approach is the range of $y^2=\dfrac{1}{{x-5}}$? And in response to that you have commented that "it's hard to say," and "it eliminates the negative numbers." thats why I mentioned that. – Shekhar Malhotra Mar 31 '20 at 18:24
  • Speaking of symmetry w.r.t. the $x$-axis, you implicitly refer to $y$, not $y^2$. – Bernard Mar 31 '20 at 19:49
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So you have to find all $y$ such that exist $x$ which satisfies $y= {1\over \sqrt{x-5}}$. Clearly, $y>0$ so $\operatorname{Range}(f)\subseteq (0,\infty)$. Now we solve this equation on $x$ and we get $$x= 5+{1\over y^2}$$ and we see, no matter what positive $y$ is, we can calculate $x$, and thus $\operatorname{Range}(f)= (0,\infty)$.

amWhy
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nonuser
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  • I had previously been your first upvote. I know perfectly well that $y$ must be positive. You're complaining to the wrong person. – amWhy Mar 31 '20 at 16:36