Question

Find the domain for the following functions.

$\rm 1] \: \: f(x) = \dfrac{ {2}^{x} }{(x - 2)} \\ \\ \rm 2] \: \: f(x) = \dfrac{ log_{e}(x - 1) }{ log_{e}(x - 2) } \\ \\ \rm 3] \: \: f(x) = log_{ |x| }( {x}^{2} + 1)$

Answer only if you know. Please answer it as soon as possible and Need Clear Explanation ​

Answer 1

1. Given,

$\longrightarrow\rm{f(x)=\dfrac{2^x}{x-2}}$

We know the denominator of a fraction cannot be zero.

$\rm{\longrightarrow x-2\neq0}$

$\rm{\longrightarrow x\neq2}$

So the domain is,

$\rm{\longrightarrow\underline{\underline{x\in[2,\ \infty)}}}$

2. Given,

$\rm{\longrightarrow f(x)=\dfrac{\log_e(x-1)}{\log_e(x-2)}}$

As the denominator of a fraction can't be zero,

$\rm{\longrightarrow \log_e(x-2)\neq0}$

$\rm{\longrightarrow x-2\neq1}$

$\rm{\longrightarrow x\neq3}$

$\mathrm{\Longrightarrow x\in}\ \mathbb{R}-\mathrm{\{3\}\quad\quad\dots(1)}$

Since the function $\rm{f(x)=\log x}$ is defined for $\rm{x>0,}$

$\rm{\longrightarrow x-1>0}$

$\rm{\longrightarrow x>1}$

$\rm{\Longrightarrow x\in(1,\ \infty)\quad\quad\dots(2)}$

And,

$\rm{\longrightarrow x-2>0}$

$\rm{\longrightarrow x>2}$

$\rm{\Longrightarrow x\in(2,\ \infty)\quad\quad\dots(3)}$

Finally, taking $\rm{(1)\land(2)\land(3),}$

$\mathrm{\longrightarrow x\in}\ [\mathbb{R}-\mathrm{\{3\}}]\cap\mathrm{(1,\ \infty)\cap(2,\ \infty)}$

$\rm{\longrightarrow\underline{\underline{x\in(2,\ \infty)-\{3\}}}}$

3. Given,

$\rm{\longrightarrow f(x)=\log_{|x|}\left(x^2+1\right)}$

Or,

$\rm{\longrightarrow f(x)=\dfrac{\log_e\left(x^2+1\right)}{\log_e|x|}}$

Since the denominator of a fraction can't be zero,

$\rm{\longrightarrow \log_e|x|\neq0}$

$\rm{\longrightarrow |x|\neq1}$

$\rm{\longrightarrow x\neq\pm1}$

$\mathrm{\Longrightarrow x\in}\ \mathbb{R}-\mathrm{\{-1,\ 1\}\quad\quad\dots(4)}$

Since the function $\rm{f(x)=\log x}$ is defined for $\rm{x>0,}$

$\rm{\longrightarrow x^2+1>0}$

$\rm{\longrightarrow x^2>-1}$

$\rm{\Longrightarrow x^2\in(-1,\ \infty)\quad\quad\dots(5.1)}$

But we know the range of the function $\rm{f(x)=x^2}$ is,

$\rm{\longrightarrow x^2\in[0,\ \infty)\quad\quad\dots(5.2)}$

Taking $(5.1)\land(5.2),$

$\rm{\longrightarrow x^2\in[0,\ \infty)}$

$\rm{\Longrightarrow x\in(-\infty,\ 0]\cup[0,\ \infty)}$

$\rm{\longrightarrow x\in}\ \mathbb{R}\quad\quad\dots\mathrm{(5)}$

And,

$\rm{\longrightarrow |x|>0}$

$\rm{\longrightarrow |x|\in(0,\ \infty)}$

$\rm{\Longrightarrow x\in(-\infty,\ 0)\cup(0,\ \infty)}$

$\mathrm{\longrightarrow x\in}\ \mathbb{R}-\mathrm{\{0\}\quad\quad\dots(6)}$

Finally, taking $\rm{(4)\land(5)\land(6),}$

$\mathrm{\Longrightarrow x\in}\ [\mathbb{R}-\mathrm{\{-1,\ 1\}}]\cap\mathbb{R}\cap[\mathbb{R}-\mathrm{\{0\}}]$

$\longrightarrow\underline{\underline{\mathrm{x\in}\ \mathbb{R}-\mathrm{\{-1,\ 0,\ 1\}}}}$

Or,

$\rm{\longrightarrow\underline{\underline{x\in(-\infty,\ -1)\cup(-1,\ 0)\cup(0,\ 1)\cup(1,\ \infty)}}}$