Question

Find domain of th function:
$\sqrt{ {x}^{2} - 3 |x| + 2 }$
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Answer 1

Given,

$\longrightarrow\sf{f(x)=\sqrt{x^2-3|x|+2}}$

The whole function is in the form of a function in x under a square root. So the function under square root should be non - negative.

Because the function $\sf{g(x)=\sqrt x}$ is defined for $\sf{x\geq0.}$

Thus,

$\longrightarrow\sf{x^2-3|x|+2\geq0}$

Case 1:- For $\sf{x\geq0,}$

$\longrightarrow\sf{x^2-3x+2\geq0}$

$\longrightarrow\sf{x^2-x-2x+2\geq0}$

$\longrightarrow\sf{x(x-1)-2(x-1)\geq0}$

$\longrightarrow\sf{(x-1)(x-2)\geq0}$

$\Longrightarrow\sf{x\in[0,\ 1]\cup[2,\ \infty)\quad[\,\because\ x\geq 0\,]}$

Case 2:- For $\sf{x\leq0,}$

$\longrightarrow\sf{x^2+3x+2\geq0}$

$\longrightarrow\sf{x^2+x+2x+2\geq0}$

$\longrightarrow\sf{x(x+1)+2(x+1)\geq0}$

$\longrightarrow\sf{(x+1)(x+2)\geq0}$

$\Longrightarrow\sf{x\in(-\infty,\ -2]\cup[-1,\ 0]\quad[\,\because\ x\leq0\,]}$

Hence the domain of $\sf{f(x)}$ is,

$\longrightarrow\sf{D=\Big[[0,\ 1]\cup[2,\ \infty)\Big]\cup\Big[(-\infty,\ -2]\cup[-1,\ 0]\Big]}$

$\Longrightarrow\sf{\underline{\underline{D=(-\infty,\ -2]\cup[-1,\ 1]\cup[2,\ \infty)}}}$

Or,

$\Longrightarrow\sf{\underline{\underline{D=\mathbb{R}-\big[(-2,\ -1)\cup(1,\ 2)\big]}}}$

Answer 2

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