Answer
$$x = - 3{\text{ and }}x = 1$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{x + 1}}{{{x^2} + 3}} \cr
& {\text{Calculate the derivative of the function using the quotient rule}} \cr
& f'\left( x \right) = \frac{{\left( {{x^2} + 3} \right)\frac{d}{{dx}}\left[ {x + 1} \right] - \left( {x + 1} \right)\frac{d}{{dx}}\left[ {{x^2} + 3} \right]}}{{{{\left( {{x^2} + 3} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{\left( {{x^2} + 3} \right)\left( 1 \right) - \left( {x + 1} \right)\left( {2x} \right)}}{{{{\left( {{x^2} + 3} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{{x^2} + 3 - 2{x^2} - 2x}}{{{{\left( {{x^2} + 3} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{ - {x^2} - 2x + 3}}{{{{\left( {{x^2} + 3} \right)}^2}}} \cr
& {\text{Set }}f'\left( x \right) = 0 \cr
& \frac{{ - {x^2} - 2x + 3}}{{{{\left( {{x^2} + 3} \right)}^2}}} = 0 \cr
& - {x^2} - 2x + 3 = 0 \cr
& {x^2} + 2x - 3 = 0 \cr
& {\text{Factoring}} \cr
& \left( {x + 3} \right)\left( {x - 1} \right) = 0 \cr
& {x_1} = - 3,{\text{ }}{x_2} = 1 \cr
& {\text{Therefore, the stationary points are:}} \cr
& x = - 3{\text{ and }}x = 1 \cr} $$
