Question

Find the domain of f(x) = $\frac{1}{\sqrt{2x^2+5x+2}}$

• A. R
• B. $(-2,\frac{-1}{2})$
• C. $(-\infty ,-2]\cup [2,\infty )$
• D. $(-\infty ,-2]\cup [\frac{-1}{2},\infty )$

Answer 1

Answer: (D)

$(-\infty ,-2]\cup [\frac{-1}{2},\infty )$

The function will have a real value only when denominator which is under a square root is $>0$
$=>2x^2+5x+2>0$
$=>2x^2+4x+x+2>0$
$=>2x(x+2)+1(x+2)>0$
$=>(2x+1)(x+2)>0$
Case 1 when both $(2x+1)(x+2)>0$
$=>x>-\frac{1}{2};or>-2$

$=>x>-\frac{1}{2}or(-\frac{1}{2},\infty )$
Case 2 when both $(2x+1)(x+2)<0$
$=>x<-\frac{1}{2};or<-2$
$=>x<-2or(\infty ,-2)$