Question

16. Find the domain of under root 9 - xsquare​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\: \sqrt{9 - {x}^{2} }$

Now,

Domain of a function f(x) is defined as set of those real values of x for which function is well defined.

So, using this definition,

$\rm :\longmapsto\: \sqrt{9 - {x}^{2} } \: is \: defined \: when \:$

$\rm :\longmapsto\:9 - {x}^{2} \geqslant 0$

$\rm :\longmapsto\: - ({x}^{2} - 9) \geqslant 0$

$\rm :\longmapsto\: {x}^{2} - 9 \leqslant 0$

$\rm :\longmapsto\: {x}^{2} - {3}^{2} \leqslant 0$

$\rm :\longmapsto\:(x - 3)(x + 3) \leqslant 0$

$\rm\implies \: - 3 \leqslant x \leqslant 3$

$\bf\implies \:x \: \in \: [ - 3, \: 3]$

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Learn to More

If a < b then

$\boxed{\tt{ (x - a)(x - b) < 0 \: \: \rm\implies \:a < x < b \: }}$

$\boxed{\tt{ (x - a)(x - b) \leqslant 0 \: \: \rm\implies \:a \leqslant x \leqslant b \: }}$

$\boxed{\tt{ (x - a)(x - b) > 0 \: \: \rm\implies \:x < a \: \: or \: \: x > b \: }}$

$\boxed{\tt{ (x - a)(x - b) \geqslant 0 \: \: \rm\implies \:x \leqslant a \: \: or \: \: x \geqslant b \: }}$

Answer 2

★find:-

$\red{DOMAIN \: \: \: OF} \\ \\ \\ \implies \: \star\pink{ \sqrt{9 - x {}^{2} } }$

solution

$\underline{DOMAIN}$

• x all posible real number value when defined by function f(x)

• $\implies \sqrt{9 - x {}^{2} }$

• we know that root pogitive because x difined real value root negative not real value

it means

$\implies \: \sqrt{9 - x {}^{2} } \geqslant 0$

do square both sides

$\implies \: (\sqrt{9 - x {}^{2} }) {}^{2} \geqslant \sqrt{0} \\ \\ \\ \implies \: 9 - x {}^{2} \geqslant0$

transposition

$\implies x {}^{2} - 9 \leqslant 0$

can we be written as

$\implies \: x {}^{2} - 3 {}^{2} \leqslant 0$

we know that

$\implies \star \pink{ a {}^{2} - b {}^{2} = ( a+b )(a - b)}$

$\implies \: (x - 3)(x + 3) \leqslant 0$

domain

$\star \red{ - 3 \leqslant x \leqslant 3}$

,

$\implies \red {x \in \:[ - 3 \: ,3]}$

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I hope it helps you