How do you find extreme points of a function?

Question

I was required to find out the range of a function(without using maxima-minima), and whatever I do, it ends up wrong. I even checked it in a calculator and its wrong. The function goes: $$f(x)=\ln(3x^2-4x+5)$$ i find the domain to be all real numbers. also, to find the range(using input restrictions ) likewise: $$3x^2-4x+5>0\\ (3x^2-4x+1)+4 > 0\\ (3x-1)(x-1) > -4$$ her again i find the range to be $\ln(-\infty,+\infty)$ but the ans is $[\sqrt{11/3},\infty)$ im stuck. how do you even get $11/3$?

im putting this again that we arent supposed to do it using maxima minima.

thanks in advance

Answer 1

We have $$3x^2-4x+5=3(x-\frac23)^2+\frac{11}{3}$$ which takes the minimum $\dfrac{11}{3}$ at $x=\dfrac{2}{3}$. Since $y=\ln x$ is increasing, therefore the range of $f(x)=\ln(3x^2-4x+5)$ is $$[\ln\frac{11}{3},\infty)$$

Answer 2

Since $\ln$ increases, we obtain: $$f(x)=\ln3+\ln\left(\left(x-\frac{2}{3}\right)^2+\frac{11}{9}\right)\geq\ln3+\ln\frac{11}{9}=\ln\frac{11}{3}.$$ The equality occurs for $x=\frac{2}{3}.$

Also, $$\lim_{x\rightarrow\infty}f(x)=+\infty$$ and $f$ is a continuous function, which says that the range it's: $$\left[\ln\frac{11}{3},+\infty\right).$$