Question

write the domain of the following function
$\sqrt{3 - 2x - {x}^{2} }$
​

Answer 1

We're given the function,

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

We've to find its domain.

We know domain is the set of all possible values of $x$ for which $f(x)$ exists.

The function $f(x)=\sqrt x$ is possible only for $x\geq0,$

$\longrightarrow3-2x-x^2\geq0}$

Multiplying both sides by -1, [note the symbol change]

$\longrightarrow x^2+2x-3\leq0}$

Factorising LHS by splitting middle term,

$\longrightarrow x^2-x+3x-3\leq0}$

$\longrightarrow x(x-1)+3(x-1)\leq0}$

$\longrightarrow (x-1)(x+3)\leq0}$

$\Longrightarrow\underline{\underline{x\in[-3,\ 1]}}$

This is the domain of the function.

Answer 2

Given ,

The function is

$\tt f(x) = \sqrt{3 - 2x - {(x)}^{2} }$

For f(x) to be defined ,

$\tt \implies 3 - 2x - {(x)}^{2} \geqslant 0$

Multiplying the inequality by -1 , we get

$\tt \implies {(x)}^{2} + 2x - 3 \leqslant 0$

$\tt \implies {(x)}^{2} - x + 3x- 3 \leqslant 0$

$\tt \implies x (x - 1) + 3(x - 1) \leqslant 0$

$\tt \implies (x + 3)(x - 1) \leqslant 0$

$\tt \implies x \: € \: [ -3 , 1]$

Therefore , domain = [-3 , 1]

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