Answer
$${\text{Relative minimum at }}\left( {0,1} \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \sqrt {{x^2} + 1} \cr
& {\text{*Calculate the first derivative}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\sqrt {{x^2} + 1} } \right] \cr
& f'\left( x \right) = \frac{{2x}}{{2\sqrt {{x^2} + 1} }} = \frac{x}{{\sqrt {{x^2} + 1} }} \cr
& {\text{Set }}f'\left( x \right) = 0 \cr
& \frac{x}{{\sqrt {{x^2} + 1} }} = 0 \cr
& x = 0 \cr
& \cr
& {\text{Calculate the second derivative}} \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {f'\left( x \right)} \right] \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{x}{{\sqrt {{x^2} + 1} }}} \right] \cr
& f''\left( x \right) = \frac{{\sqrt {{x^2} + 1} - x\left( {\frac{x}{{\sqrt {{x^2} + 1} }}} \right)}}{{{x^2} + 1}} \cr
& f''\left( x \right) = \frac{{{x^2} + 1 - {x^2}}}{{{{\left( {{x^2} + 1} \right)}^{3/2}}}} \cr
& f''\left( x \right) = \frac{1}{{{{\left( {{x^2} + 1} \right)}^{3/2}}}} \cr
& {\text{Evaluate the second derivative at }}x = 0 \cr
& f''\left( 0 \right) = \frac{1}{{{{\left( {{0^2} + 1} \right)}^{3/2}}}} > 0 \cr
& {\text{Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr
& f\left( x \right){\text{ has a relative minimum at }}\left( {0,f\left( 0 \right)} \right) \cr
& f\left( 0 \right) = \sqrt {{0^2} + 1} \cr
& {\text{Relative minimum at }}\left( {0,1} \right) \cr} $$
