Question

Find domain and range of the real function f(x)=
3/1-x2​

Answer 1

$\rm \: f(x) = \dfrac{3}{1 - {x}^{2} } \\ \\ \rm \longrightarrow1 - {x}^{2} \neq0 \\ \\ \rm \longrightarrow {x}^{2} \neq1\\ \\ \rm \longrightarrow {x} \neq \: \pm1$

Therefore, domain of given function = x ∈ R - {±1}

Now, we will find out range of the function.

Let f(x) be y.

$\rm \longrightarrow\: y = \dfrac{3}{1 - {x}^{2} } \\ \\ \rm \longrightarrow\: y ({1 - {x}^{2} } )= {3} \\ \\ \rm \longrightarrow\: ({1 - {x}^{2} } )= \frac{3}{y} \\ \\ \rm \longrightarrow\: {- {x}^{2} } = \frac{3}{y} - 1\\ \\ \rm \longrightarrow\: { {x}^{2} } = 1 - \frac{3}{y} \\ \\ \rm \longrightarrow\: { {x} } = \sqrt{1 - \frac{3}{y} } \\ \\ \rm \rightarrow\: y \neq0 \\ \\ \rm \rightarrow1 - \frac{3}{y} \geqslant 0\\ \\ \\ \rm \rightarrow1 \geqslant \frac{3}{y} \\ \: \\ \rm \rightarrow \: y \geqslant {3} \\ \\ \\ \rm \: \therefore \: range = y \in \: [3,\infty)$