Question

Find the domain and range of $f\left(x\right)=\sqrt{3x-5}$.

Answer 1

$f\left(x\right)=\sqrt{3x-5}$

$3x-5\ge 0$

$\Rightarrow$ $3x\ge 5$

$\Rightarrow$ $x\ge \frac{5}{3}$ ---- ( 1 )

$\therefore$ Domain $=[\frac{5}{3},\infty )$

For range :

$y=\sqrt{3x-5}$

$\Rightarrow$ $y^2=3x-5$

$\Rightarrow$ $3x=5+y^2$

$\Rightarrow$ $x=\frac{5+y^2}{3}$

But $x\ge \frac{5}{3}$ [ From ( 1 ) ]

$\therefore$ $\frac{5}{3}+\frac{y^2}{3}\ge \frac{5}{3}$

$\Rightarrow$ $\frac{y^2}{3}\ge 0$

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

$\Rightarrow$ $y\ge 0$

$\therefore$ Rage $=$ The real numbers $=[0,\infty )$