Answer
$$\eqalign{
& {\text{critical points: }}\left( { \pm \frac{1}{2},0} \right) \cr
& {\text{inflection points: }}none \cr} $$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = \sqrt {4{x^2} - 1} \cr
& {\text{Differentiate }}f'\left( x \right) \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\sqrt {4{x^2} - 1} } \right] \cr
& f'\left( x \right) = \frac{{8x}}{{2\sqrt {4{x^2} - 1} }} \cr
& f'\left( x \right) = \frac{{4x}}{{\sqrt {4{x^2} - 1} }} \cr
& {\text{Calculate the critical points set the derivative equal to 0}} \cr
& \frac{{4x}}{{\sqrt {4{x^2} - 1} }} = 0 \cr
& 4x = 0 \cr
& x = 0 \cr
& {\text{This value is not in the domain}} \cr
& {\text{The derivative is not defined when }}4{x^2} - 1 = 0,{\text{ then}} \cr
& 4{x^2} - 1 = 0 \cr
& {x^2} = \frac{1}{4} \cr
& x = \pm \frac{1}{2} \cr
& {\text{Calculate }}f\left( { \pm \frac{1}{2}} \right) \cr
& f\left( {\frac{1}{2}} \right) = 0,{\text{ }}f\left( { - \frac{1}{2}} \right) = 0 \cr
& {\text{Therefore, the critical points are:}} \cr
& \left( { \pm \frac{1}{2},0} \right) \cr
& \cr
& {\text{Calculate the second derivative}} \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{4x}}{{\sqrt {4{x^2} - 1} }}} \right] \cr
& f''\left( x \right) = \frac{{\left( {\sqrt {4{x^2} - 1} } \right)\left( 4 \right) - 4x\left( {\frac{{4x}}{{\sqrt {4{x^2} - 1} }}} \right)}}{{{{\left( {\sqrt {4{x^2} - 1} } \right)}^2}}} \cr
& f''\left( x \right) = \frac{{4\sqrt {4{x^2} - 1} - \frac{{16{x^2}}}{{\sqrt {4{x^2} - 1} }}}}{{4{x^2} - 1}} \cr
& f''\left( x \right) = \frac{{4\left( {4{x^2} - 1} \right) - 16{x^2}}}{{{{\left( {4{x^2} - 1} \right)}^{3/2}}}} \cr
& f''\left( x \right) = \frac{{16{x^2} - 4 - 16{x^2}}}{{{{\left( {4{x^2} - 1} \right)}^{3/2}}}} \cr
& f''\left( x \right) = - \frac{4}{{{{\left( {4{x^2} - 1} \right)}^{3/2}}}} \cr
& {\text{Set }}f''\left( x \right) = 0 \cr
& {\text{The second derivative is never 0, then there are no inflection}} \cr
& {\text{points}}{\text{.}} \cr} $$
