Answer
$${\text{No relative extrema}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{x}{{x + 2}} \cr
& {\text{Calculate the derivative of }}f\left( x \right){\text{ by using the quotient rule}} \cr
& f'\left( x \right) = \frac{{\left( {x + 2} \right)\left( 1 \right) - x\left( 1 \right)}}{{{{\left( {x + 2} \right)}^2}}} \cr
& {\text{Find the critical points, set }}f'\left( x \right) = 0 \cr
& f'\left( x \right) = \frac{{x + 2 - x}}{{{{\left( {x + 2} \right)}^2}}} \cr
& f'\left( x \right) = \frac{2}{{{{\left( {x + 2} \right)}^2}}} \cr
& {\text{Find the critical points, set }}f'\left( x \right) = 0 \cr
& \frac{2}{{{{\left( {x + 2} \right)}^2}}} = 0 \cr
& {\text{There are no real values for which }}f'\left( x \right) = 0 \cr
& {\text{Then, the function does not have relative extrema}}{\text{.}} \cr} $$
