I'm having trouble to find the range of more complicated functions such as

$ f(x) = \frac{1}{\sqrt{x + 1}} $

How one should proceed to tackled down these functions, specially the ones involving roots and quotients, without using tools from calculus? (limits, derivatives, etc.)

Thanks in advance.

  • 1
    $\begingroup$ Hint: Consider $\sqrt {x+1}$ where $x \lt -1$. Also consider what happens to $\frac 1{\sqrt{x+1}}$ when $x=-1$. $\endgroup$ Commented Nov 15, 2018 at 2:24

1 Answer 1


We proceed using inequalities. For high school students it is more comprehensible if we formally use comparison with $\infty.$

  • For the given function, start from its domain $$\big(0<x+1<\infty\big) \Longrightarrow \big(0<\sqrt{x+1}<\infty\big) \Longrightarrow \big(\infty>{1\over{\sqrt{x+1}}}>0\big), $$ so the range is $(0,\infty).$

  • It becomes more interesting if e.g. $\;g(x) = \frac{3}{\sqrt{(x + 1)}+2}$

$$\begin{aligned}\big(0<x+1<\infty\big) \Longrightarrow &\;0<\sqrt{x+1}<\infty\\ \Longrightarrow &\;2<\sqrt{x+1}+2<\infty\\\Longrightarrow &\;{1\over 2}>{1\over{\sqrt{(x+1)}+2}}>0\\\Longrightarrow &\;{3\over 2}>g(x)>0\end{aligned}$$ thus the range is $(0,{3\over 2}).$

  • Consider now $\;h(x) = \frac{3}{\sqrt{(x + 1)}-2},$ its domain is $x \in (-1,3) \cup (3,\infty).$ For the sign of $h(x),$ we have to consider the intervals separately. The method still works. $$\begin{aligned}\big(0<x+1<4\big) \Longrightarrow &\;0<\sqrt{x+1}<2\\ \Longrightarrow &\;-2<\sqrt{x+1}-2<0\\\Longrightarrow &\;{-1\over 2}>{1\over{\sqrt{(x+1)}-2}}>-\infty\\\Longrightarrow &\;{-3\over 2}>h(x)>-\infty\end{aligned}$$ $$\begin{aligned}\big(4<x+1<\infty\big) \Longrightarrow &\;2<\sqrt{x+1}<\infty\\ \Longrightarrow &\;0<\sqrt{x+1}-2<\infty\\\Longrightarrow &\;\infty>{1\over{\sqrt{(x+1)}-2}}>0\\\Longrightarrow &\;\infty>h(x)>0\end{aligned}$$
  • $\begingroup$ Interesting method, but how does it work when you have a $ - 2 $ instead of a $ + 2 $ in the denominator? $\endgroup$ Commented Nov 20, 2018 at 13:56
  • 1
    $\begingroup$ Edited for a function $h.$ $\endgroup$
    – user376343
    Commented Nov 22, 2018 at 9:16

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