The range of the function f (x) = log_e (3x² – 4x + 5) is

f (x) is defined if 3x² – 4x + 5 > 0
$\Rightarrow 3\left[{\mathrm{x}}^2-\frac{4}{3}\mathrm{x}+\frac{5}{3}\right]>0\Rightarrow 3\left[{\left(\mathrm{x}-\frac{2}{3}\right)}^2+\frac{11}{9}\right]>0$
which is true for all real x.
$\therefore$ Domain ( f ) = (– ∞, ∞)
$$\begin{gathered} \mathrm{y}={\log }_{\mathrm{e}}\left(3{\mathrm{x}}^2-4\mathrm{x}+5\right)\Rightarrow {\mathrm{e}}^{\mathrm{y}}=3{\mathrm{x}}^2-4\mathrm{x}+5 \\ \Rightarrow 3{\mathrm{x}}^2-4\mathrm{x}+\left(5-{\mathrm{e}}^{\mathrm{y}}\right)=0 \end{gathered}$$
For x to be real,
$$\begin{gathered} 16-12\left(5-{\mathrm{e}}^{\mathrm{y}}\right)\ge 0\Rightarrow 12{\mathrm{e}}^{\mathrm{y}}\ge 44\Rightarrow {\mathrm{e}}^{\mathrm{y}}\ge \frac{11}{3} \\ \begin{array}{l}\Rightarrow \mathrm{y}\ge {\log }_{\mathrm{e}}\frac{11}{3}\\ \therefore \text{ Range of }\mathrm{f}=\left[{\log }_{\mathrm{e}}\frac{11}{3},\mathrm{\infty }\right)\end{array} \end{gathered}$$