Author Topic: Maximal Domains - for one to one functions (Read 1149 times) Tweet Share

zhe0001 · « **on:** June 05, 2013, 06:12:59 pm »

Confused about a maths question,
it asks to find two maximal domains that allow the function to become a one to one function
The function,
$1/(1+x^2)$

I notice that it's a truncus function and that the $(1+x^2)$ means that the vertical asymptote will be $x=-1$
I thought the maximal domains would be (-infinite, -1] and [-1. infinite)

But apparently not, the answer is (-infinite, 0] and [0, infinite)

Please tell me where I went wrong.
Thanks in advance!

Phy124 · « **Reply #1 on:** June 05, 2013, 06:45:04 pm »

I think you are confusing $\frac{1}{1+x^2}$ with either of $\frac{1}{x^2-1}$ or $\frac{1}{1-x^2}$ .

$\frac{1}{1+x^2}$ does not have any vertical asymptotes as $1+x^2>0$ for $x \in \mathbb{R}$

zhe0001 · « **Reply #2 on:** June 05, 2013, 07:24:35 pm »

Quote from: 2/cos(c) on June 05, 2013, 06:45:04 pm

I think you are confusing $\frac{1}{1+x^2}$ with either of $\frac{1}{x^2-1}$ or $\frac{1}{1-x^2}$ .

$\frac{1}{1+x^2}$ does not have any vertical asymptotes as $1+x^2>0$ for $x \in \mathbb{R}$

Understood. Thanks!

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Author Topic: Maximal Domains - for one to one functions (Read 1149 times) Tweet Share

zhe0001

Maximal Domains - for one to one functions

Phy124

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Re: Maximal Domains - for one to one functions

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