Question

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

• A. $[-\infty ,\sqrt{\frac{11}{3}}]$
• B. $(-\infty ,\sqrt{\frac{11}{3}})$
• C. $[\sqrt{\frac{11}{3}},\infty )$
• D. $(\sqrt{\frac{11}{3}},\infty )$

Answer 1

The correct option is C $[\sqrt{\frac{11}{3}},\infty )$

$f\left(x\right)$ is defined if $3x^2-4x+5>0$

$\Rightarrow 3[x^2-\frac{4}{3}x+\frac{5}{3}]>0\Rightarrow 3[{(x-\frac{2}{3})}^2+\frac{11}{9}]>0$

which is true for all real $x$
$\therefore$ Domain $f\left(x\right)=(-\infty ,\infty )$
Let $y=\sqrt{3x^2-4x+5}\Rightarrow 3x^2-4x+(5-y^2)=0$

For $x$ to be real
$16-12(5-y^2)\ge 0\Rightarrow y\ge \frac{11}{3}$

$\therefore$ Range of $f\left(x\right)=[\sqrt{\frac{11}{3}},\infty )$