Question

The domain of the function f(x)=√9 − x to the power
2 is

Please do help me fast. I will mark u brainliest.

Answer 1

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

Given function is

$\rm :\longmapsto\:f(x) = \sqrt{9 - {x}^{2} }$

We know,

Domain of a function is defined as set of those values of x, for which function is well defined.

So,

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

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

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

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

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

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

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

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

So, option a) is correct

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information

$\boxed{\tt{ |x| < y \: \rm \implies\: - y < x < y \: }}$

$\boxed{\tt{ |x| \leqslant y \: \rm \implies\: - y \leqslant x \leqslant y \: }}$

$\boxed{\tt{ |x| > y \: \rm \implies\: \: x < - y \: \: or \: \: x > y \: }}$

$\boxed{\tt{ |x| \geqslant y \: \rm \implies\: \: x \leqslant - y \: \: or \: \: x \geqslant y \: }}$

$\boxed{\tt{ |x - a| < y \: \rm \implies\:a - y < x <a + y \: }}$

$\boxed{\tt{ |x - a| \leqslant y \: \rm \implies\:a - y \leqslant x \leqslant a + y \: }}$