Answer
$$\eqalign{
& {\text{Inflection point }}\left( {0,0} \right) \cr
& {\text{Vertical asymptotes: none}} \cr
& {\text{Horizontal asymptotes at }}y = - \frac{3}{2}{\text{ and }}y = \frac{3}{2} \cr
& {\text{No relative extrema}} \cr
& {\text{Domain}}:\left( { - \infty ,\infty } \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{3x}}{{\sqrt {4{x^2} + 1} }} \cr
& 4{x^2} + 1{\text{ is always positive and }} > {\text{0, then the domain of the }} \cr
& {\text{function is }}D:\left( { - \infty ,\infty } \right) \cr
& \cr
& {\text{*Differentiating}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{3x}}{{\sqrt {4{x^2} + 1} }}} \right] \cr
& f'\left( x \right) = \frac{{3\sqrt {4{x^2} + 1} - 3x\left( {\frac{{8x}}{{2\sqrt {4{x^2} + 1} }}} \right)}}{{4{x^2} + 1}} \cr
& f'\left( x \right) = \frac{{6\left( {4{x^2} + 1} \right) - 3x\left( {8x} \right)}}{{2{{\left( {4{x^2} + 1} \right)}^{3/2}}}} \cr
& f'\left( x \right) = \frac{{24{x^2} + 6 - 24{x^2}}}{{2{{\left( {4{x^2} + 1} \right)}^{3/2}}}} \cr
& f'\left( x \right) = \frac{6}{{2{{\left( {4{x^2} + 1} \right)}^{3/2}}}} \cr
& {\text{Set }}f'\left( x \right) = 0 \cr
& \frac{3}{{{{\left( {4{x^2} + 1} \right)}^{3/2}}}} = 0 \cr
& {\text{There are no solution}}{\text{.}} \cr
& {\text{Then there are no relative extrema}}{\text{.}} \cr
& \cr
& {\text{*Find the second derivative}} \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{3}{{{{\left( {4{x^2} + 1} \right)}^{3/2}}}}} \right] \cr
& {\text{Using a CAS we obtain}} \cr
& f''\left( x \right) = \frac{9}{2}{\left( {4{x^2} + 1} \right)^{1/2}}\left( {8x} \right) \cr
& f''\left( x \right) = 36x{\left( {4{x^2} + 1} \right)^{1/2}} \cr
& {\text{Set }}f''\left( x \right) = 0 \cr
& 36x = 0 \cr
& x = 0 \cr
& {\text{For }}\left( { - \infty ,0} \right) \to f''\left( { - 1} \right) > 0 \cr
& {\text{For }}\left( {0,\infty } \right) \to f''\left( 1 \right) < 0 \cr
& {\text{Then there is an inflection point at }}x = 0 \cr
& f\left( 0 \right) = 0 \to {\text{Inflection point }}\left( {0,0} \right) \cr
& \cr
& {\text{There are no vertical asymptotes because the denominator}} \cr
& {\text{is never 0}}{\text{.}} \cr
& \cr
& *{\text{Find the horizontal asymptotes}} \cr
& \mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \frac{{3x}}{{\sqrt {4{x^2} + 1} }} = \frac{3}{2} \cr
& \mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to - \infty } \frac{{3x}}{{\sqrt {4{x^2} + 1} }} = - \frac{3}{2} \cr
& {\text{Horizontal asymptotes at }}y = - \frac{3}{2}{\text{ and }}y = \frac{3}{2} \cr
& \cr
\cr} $$