The range of the function $\mathrm{f}\left(\mathrm{x}\right)=\sqrt{3{\mathrm{x}}^2-4\mathrm{x}+5} \mathrm{is}$

f (x) is defined if 3x2 – 4x + 5 ≥ 0
$\Rightarrow 3\left[{\mathrm{x}}^2-\frac{4}{3}\mathrm{x}+\frac{5}{3}\right]\ge 0\Rightarrow 3\left[{\left(\mathrm{x}-\frac{2}{3}\right)}^2+\frac{11}{9}\right]\ge 0$
which is true for all real x
$\therefore$ Domain ( f ) = (– ∞, ∞)
$\text{ Let }\mathrm{y}=\sqrt{3{\mathrm{x}}^2-4\mathrm{x}+5}$
⇒ y² = 3x² – 4x + 5 i.e. 3x² – 4x + (5 – y²) = 0
For x to be real, 16 – 12 (5 – y²) ≥ 0 $\Rightarrow \mathrm{y}\ge \sqrt{\frac{11}{3}}$
$\therefore \text{ Range of }\mathrm{y}=\left[\sqrt{\frac{11}{3}},\mathrm{\infty }\right)$