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$$\frac{\left(-2 + \frac{\left(2 - \frac{2 x - 3}{x + 1}\right) \left(2 x - 3\right)}{\left(1 - \frac{\left(2 x - 3\right)^{2}}{\left(x + 1\right)^{2}}\right) \left(x + 1\right)}\right) \left(2 - \frac{2 x - 3}{x + 1}\right)}{\sqrt{1 - \frac{\left(2 x - 3\right)^{2}}{\left(x + 1\right)^{2}}} \left(x + 1\right)^{2}} = 0$$
Solve this equationThe roots of this equation
$$x_{1} = \frac{3}{2} - \frac{5 \sqrt{3}}{6}$$
$$x_{2} = \frac{5 \sqrt{3}}{6} + \frac{3}{2}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -1$$
$$\lim_{x \to -1^-}\left(\frac{\left(-2 + \frac{\left(2 - \frac{2 x - 3}{x + 1}\right) \left(2 x - 3\right)}{\left(1 - \frac{\left(2 x - 3\right)^{2}}{\left(x + 1\right)^{2}}\right) \left(x + 1\right)}\right) \left(2 - \frac{2 x - 3}{x + 1}\right)}{\sqrt{1 - \frac{\left(2 x - 3\right)^{2}}{\left(x + 1\right)^{2}}} \left(x + 1\right)^{2}}\right) = - \infty i$$
$$\lim_{x \to -1^+}\left(\frac{\left(-2 + \frac{\left(2 - \frac{2 x - 3}{x + 1}\right) \left(2 x - 3\right)}{\left(1 - \frac{\left(2 x - 3\right)^{2}}{\left(x + 1\right)^{2}}\right) \left(x + 1\right)}\right) \left(2 - \frac{2 x - 3}{x + 1}\right)}{\sqrt{1 - \frac{\left(2 x - 3\right)^{2}}{\left(x + 1\right)^{2}}} \left(x + 1\right)^{2}}\right) = \infty i$$
- the limits are not equal, so
$$x_{1} = -1$$
- 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
$$\left[\frac{5 \sqrt{3}}{6} + \frac{3}{2}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, \frac{5 \sqrt{3}}{6} + \frac{3}{2}\right]$$