Question

Find the domain :

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

• A. $[0,\frac{1}{2}]$
• B. $R-${$-2$}
• C. $[-\frac{1}{2},\frac{1}{2}]$
• D. $[-1,1]$

Answer 1

The correct option is B $R-${$-2$}

We have, $f\left(x\right)=\frac{1}{x+2}$

Clearly, $f\left(x\right)$ assumes real values for all real values of $x$ except for the values of $x$ satisfying $x+2=0⟹x=-2$.

Hence, Domain $\left(f\right)=R-\{-2\}.$