# There is a general method for finding ranges of function without resorting to calculus?

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.)

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

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 $$\big(0{1\over{\sqrt{x+1}}}>0\big),$$ 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{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{1\over{\sqrt{(x+1)}-2}}>-\infty\\\Longrightarrow &\;{-3\over 2}>h(x)>-\infty\end{aligned} \begin{aligned}\big(4{1\over{\sqrt{(x+1)}-2}}>0\\\Longrightarrow &\;\infty>h(x)>0\end{aligned}
• Interesting method, but how does it work when you have a $- 2$ instead of a $+ 2$ in the denominator? Commented Nov 20, 2018 at 13:56
• Edited for a function $h.$ Commented Nov 22, 2018 at 9:16