Let's find the inflection points, we'll need to solve the equation for this
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = $$
the second derivative$$\frac{\frac{2 \sqrt{x - 2}}{\left(x + 2\right)^{2}} - \frac{1}{\sqrt{x - 2} \left(x + 2\right)} - \frac{1}{4 \left(x - 2\right)^{\frac{3}{2}}}}{x + 2} = 0$$
Solve this equationThe roots of this equation
$$x_{1} = 6 - \frac{8 \sqrt{3}}{3}$$
$$x_{2} = \frac{8 \sqrt{3}}{3} + 6$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -2$$
$$\lim_{x \to -2^-}\left(\frac{\frac{2 \sqrt{x - 2}}{\left(x + 2\right)^{2}} - \frac{1}{\sqrt{x - 2} \left(x + 2\right)} - \frac{1}{4 \left(x - 2\right)^{\frac{3}{2}}}}{x + 2}\right) = - \infty i$$
$$\lim_{x \to -2^+}\left(\frac{\frac{2 \sqrt{x - 2}}{\left(x + 2\right)^{2}} - \frac{1}{\sqrt{x - 2} \left(x + 2\right)} - \frac{1}{4 \left(x - 2\right)^{\frac{3}{2}}}}{x + 2}\right) = \infty i$$
- the limits are not equal, so
$$x_{1} = -2$$
- is an inflection point
Сonvexity and concavity intervals:Let’s find the intervals where the function is convex or concave, for this look at the behaviour of the function at the inflection points:
Concave at the intervals
$$\left[\frac{8 \sqrt{3}}{3} + 6, \infty\right)$$
Convex at the intervals
$$\left(-\infty, \frac{8 \sqrt{3}}{3} + 6\right]$$