Heelp with stationery points

[nil101]

Find the coordinates of the stationary points on the curve
\(\displaystyle \L

y = 2x + \frac{3}{x}\)
leaving your answer in surd form.


\(\displaystyle \L
\frac{{dy}}{{dx}} = 2 - 3x^{ - 2} = 0\)
\(\displaystyle x = \pm \sqrt {\frac{3}
{2}}\)
The stationary points are:
\(\displaystyle \begin{array}{l}
A = \left( {\sqrt {\frac{3}{2}} ,2\sqrt 6 } \right) \\
B = \left( { - \sqrt {\frac{3}{2}} , - 2\sqrt 6 } \right) \\
\end{array}\)

b) Using the second derivative method determine the nature of the stationary points.
\(\displaystyle \L
\frac{{d^2 y}}{{dx^2 }} = 6x^{ - 3}\)

substituting \(\displaystyle x = \pm\sqrt {\frac{3}{2}}\) I make both the 2nd derivatives equal to zero so i dont know where i'm going wrong or what the nature of the stationary points is??

Can you help please?

 

[pka]

“Using the second derivative method, determine the nature of the stationary points.”
This means to find the ‘sign’ of the second derivative at the point.
A plus means a minimum; a minus means a maximum.
\(\displaystyle \L
6\left( { \pm \sqrt {\frac{3}{2}} } \right)^{ - 3} = 6\left( { \pm \sqrt {\frac{8}{{27}}} } \right) = \pm 4\sqrt {\frac{2}{3}}\).