Answer
$${\text{Relative minimum of 0 at }}x = 0$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \ln \left( {1 + {x^2}} \right) \cr
& {\text{Differentiate both sides with respect to }}x \cr
& f'\left( x \right) = \frac{{2x}}{{1 + {x^2}}} \cr
& {\text{Set }}f'\left( x \right) = 0 \cr
& \frac{{2x}}{{1 + {x^2}}} = 0 \cr
& x = 0 \cr
& {\text{Calculate the second derivative}} \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{2x}}{{1 + {x^2}}}} \right] \cr
& f''\left( x \right) = \frac{{2\left( {1 + {x^2}} \right) - 2x\left( {2x} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr
& f''\left( x \right) = \frac{{2 + 2{x^2} - 4{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr
& f''\left( x \right) = \frac{{2 - 2{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr
& {\text{Evaluate }}f''\left( 0 \right) \cr
& f''\left( 0 \right) = \frac{{2 - 2{{\left( 0 \right)}^2}}}{{{{\left( {1 + {{\left( 0 \right)}^2}} \right)}^2}}} = 2 \cr
& f''\left( 0 \right) > 0,{\text{ Therefore, there is a relative minimum at }}x = 0 \cr
& f\left( 0 \right) = \ln \left( {1 + {{\left( 0 \right)}^2}} \right) \cr
& f\left( 0 \right) = 0 \cr
& \cr
& {\text{Relative minimum of 0 at }}x = 0 \cr} $$
