Question

the domain of f(x) = x^2/x^2 + 3​

Answer 1

Step-by-step explanation:

Let

$y = \frac{ {x}^{2} }{ {x}^{2} + 3}$

$\implies \: y ( {x}^{2} + 3)= {x}^{2}$

$\implies \: y {x}^{2} + 3y= {x}^{2}$

$\implies \: y {x}^{2} - {x}^{2} + 3y= 0$

$\implies \: (y - 1) {x}^{2} + 3y= 0$

$\implies \: {x}^{2} = \frac{3y}{1 - y}$

$\implies \: x = \sqrt{\frac{3y}{1 - y} }$

Now,

$\implies \: \frac{3y}{1 - y} \geqslant 0$

$\implies \: \frac{y}{1 - y} \geqslant 0$

$\implies \: \frac{y}{ y - 1} \leqslant 0$

$\implies\: y\in[0,1)$