Question

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

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

Answer 1

The correct option is D $(-\infty ,-2)\cup (\frac{-1}{2},\infty )$

Given $f\left(x\right)=\frac{1}{\sqrt{2x^2+5x+2}}$

$f\left(x\right)=\frac{1}{\sqrt{2x^2+4x+x+2}}$

$=\frac{1}{\sqrt{2x(x+2)+1(x+2)}}$

$=\frac{1}{\sqrt{(2x+1)(x+2)}}$

Domain is the set of possible $x$ values for which $f\left(x\right)$ is defined.

Hence for $f\left(x\right)$ to be defined, either $2x+1>0$ and $x+2>0$

or $2x+1<0$ and $x+2<0$.

In the first case, $2x+1>0⟹x>-\frac{1}{2}$ and $x+2>0⟹x>-2$.

Hence $x>-\frac{1}{2}$ ----(1)

In the second case, $2x+1<0⟹x<-\frac{1}{2}$ and $x+2<0⟹x<-2$

Hence $x<-2$ ----(2)

From (1) and (2), domain of $f\left(x\right)$ is $(-\infty ,-2)\cup (\frac{-1}{2},\infty )$