Question

Find the domain and range of the following functions:
(i) $f\left(x\right)=x^2$
(ii) $f\left(x\right)=\sqrt{(x-1)(3-x)}$

Answer 1

Solution :-

$\left(i\right)f\left(x\right)=x^2$

Domian $=R$ Range $=R^{+}$

$\left(ii\right)f\left(x\right)=\sqrt{(x-1)(3-x)}$

Domian : $(x-1)(3-x)\ge 0\Rightarrow (x-1)(x-3)\le 0$

Domain : $xϵ[1,3]$

Range :-

Let $y=\sqrt{(x-1)(3-x)}\Rightarrow y^2=(x-1)(3-x)$

$\Rightarrow y^2=3x-x^2-3+x\Rightarrow -x^2+4x-3$

$\Rightarrow x^2-4x+3+y^2=0$

$\Rightarrow x=\frac{4\pm \sqrt{16-4(3+y^2)}}{2}$

Now above equation is defined for

$16-4(3+y^2)=0$

$\Rightarrow 4\ge (3+y^2)$

$\Rightarrow 1\ge y^2$

$\Rightarrow 1-y^2\ge 0$

$\Rightarrow (1-y)(1+y)\ge 0$

Range : $yϵ(-\infty ,1]\cup [1,\infty )$