Question

find the domain y = √(1 -x) (x+3)​

Answer 1

Answer:

x = (y^2 +x^2-3) / -2

Step-by-step explanation:

y = √ (1-x) (x+3)

or, y = √ x+3 -x^2 -3x

or, y = √ -x^2 -2x +3

or, y^2 = -x^2 -2x +3

or, y^2+x^2-3 = -2x

or, x = (y^2 +x^2-3) / -2

Answer 2

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

Given function is

$\rm :\longmapsto\:y = \sqrt{(1 - x)(x + 3)}$

Now, we know

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

So, by using Definition of domain,

$\rm :\longmapsto\:y = \sqrt{(1 - x)(x + 3)} \: is \: defined \: for \: real \: values \: when$

$\rm :\longmapsto\:(1 - x)(x + 3) \geqslant 0$

$\rm :\longmapsto\: - (x - 1)(x + 3) \geqslant 0$

$\rm :\longmapsto\: (x - 1)(x + 3) \leqslant 0$

$\bf\implies \: - 3 \leqslant x \leqslant 1$

OR

$\bf\implies \:x \in \: [ - 3, \: 1]$

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MORE TO KNOW

If a and b are two real numbers such that a < b, then

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

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

$\boxed{\tt{ (x - a)(x - b) > 0 \: \: \rm\implies \: \: x \in \: ( - \infty ,a) \cup (b, \infty ) \: }}$

$\boxed{\tt{ (x - a)(x - b) \geqslant 0 \: \: \rm\implies \: \: x \in \: ( - \infty ,a] \cup [b, \infty ) \: }}$

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

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

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

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