Question

Find the domain of definition and the ranges of the following function:
$y=\log (3x^2-4x+5).$

Answer 1

$y=log({3x}^2-4x+5)$

For $y$ to be real and defined,

${3x}^2-4x+5>0$

${3x}^2-4x+5>0\forall x\in R$ (real numbers)

Therefore domain of $y$ is $x\in R$

Minimum value of ${3x}^2-4x+5$is at $x=\frac{-(-4)}{2\left(3\right)}=\frac{2}{3}$

At $x=\frac{2}{3},{3x}^2-4x+5=\frac{11}{5}>1$

${3x}^2-4x+5\in [\frac{11}{3},\infty )\forall x\in R$

Therefore, range of $y$ is $\left[log\right(\frac{11}{3}),\infty )$