Question

Find the domain of definition and the range of the following function:
$y=\sqrt{3x^2-4x+5}$

Answer 1

$y=\sqrt{3x^2-4x+5}$

For definition of square roots, $3x^2-4x+5\ge 0$

$\Rightarrow 3x^2-4x+5\ge 0$

$\Rightarrow x-\frac{4\pm \sqrt{16-\left(15\right)\left(4\right)}}{2\left(3\right)}\ge 0$

$\Rightarrow$ here we see, $D<0$

and $a=3>0,\Rightarrow f\left(x\right)=3x^2-4x+5\ge 0\forall x\in R$

So, domain is $x\in R$ and range $x\in R^{+}-O$