Question Number 31811 by NECx last updated on 15/Mar/18
![find the domain of f(x)=((2x+7)/([2−x^2 ]))](https://www.tinkutara.com/question/Q31811.png)
$${find}\:{the}\:{domain}\:{of} \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}{x}+\mathrm{7}}{\left[\mathrm{2}−{x}^{\mathrm{2}} \right]} \\ $$
Commented by Tinkutara last updated on 15/Mar/18
![Domain=R−{[−(√2),−1)∪(1,(√2)]}](https://www.tinkutara.com/question/Q31814.png)
$${Domain}={R}−\left\{\left[−\sqrt{\mathrm{2}},−\mathrm{1}\right)\cup\left(\mathrm{1},\sqrt{\mathrm{2}}\right]\right\} \\ $$
Commented by NECx last updated on 15/Mar/18
$${please}\:{show}\:{workings} \\ $$
Answered by Tinkutara last updated on 16/Mar/18
![First consider [2−x^2 ]=0 0≤2−x^2 <1 1<x^2 ≤2 x∈[−(√2),−1)∪(1,(√2)] ∴ Domain=R−{[−(√2),−1)∪(1,(√2)]}](https://www.tinkutara.com/question/Q31861.png)
$${First}\:{consider}\:\left[\mathrm{2}−{x}^{\mathrm{2}} \right]=\mathrm{0} \\ $$$$\mathrm{0}\leqslant\mathrm{2}−{x}^{\mathrm{2}} <\mathrm{1} \\ $$$$\mathrm{1}<{x}^{\mathrm{2}} \leqslant\mathrm{2} \\ $$$${x}\in\left[−\sqrt{\mathrm{2}},−\mathrm{1}\right)\cup\left(\mathrm{1},\sqrt{\mathrm{2}}\right] \\ $$$$\therefore\:{Domain}={R}−\left\{\left[−\sqrt{\mathrm{2}},−\mathrm{1}\right)\cup\left(\mathrm{1},\sqrt{\mathrm{2}}\right]\right\} \\ $$