Can I square both sides while calculating the Range of a function

Question

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?

Answer 1

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, $$\text{Im}(f)=(0,\infty ).$$

Answer 2

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 $\operatorname{Range}(f)= (0,\infty)$.

Answer 3

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