Question

The range of $f\left(x\right)=\sqrt{x^2+4x+29}$ is

• A. $(-\infty ,\infty )$
• B. $(0,\infty )$
• C. $[5,\infty )$
• D. $(0,5)$

Answer 1

Answer: (C)

$[5,\infty )$

$f\left(x\right)=\sqrt{x^2+4x+29}$
$=\sqrt{x^2+4x+4+25}$
$=\sqrt{(x+2{)}^2+5^2}$
$\because (x+2{)}^2⩾0$
$\therefore 5⩽f\left(x\right)<\infty$