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$$\frac{\frac{3 \sqrt{x - 1}}{\left(x + 1\right)^{2}} - \frac{2}{\sqrt{x - 1} \left(x + 1\right)} - \frac{1}{\left(x - 1\right)^{\frac{3}{2}}}}{4 \sqrt{x + 1}} = 0$$
Solve this equationThe roots of this equation
$$x_{1} = \frac{1}{2}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = -1$$
$$\lim_{x \to -1^-}\left(\frac{\frac{3 \sqrt{x - 1}}{\left(x + 1\right)^{2}} - \frac{2}{\sqrt{x - 1} \left(x + 1\right)} - \frac{1}{\left(x - 1\right)^{\frac{3}{2}}}}{4 \sqrt{x + 1}}\right) = \infty$$
$$\lim_{x \to -1^+}\left(\frac{\frac{3 \sqrt{x - 1}}{\left(x + 1\right)^{2}} - \frac{2}{\sqrt{x - 1} \left(x + 1\right)} - \frac{1}{\left(x - 1\right)^{\frac{3}{2}}}}{4 \sqrt{x + 1}}\right) = \infty i$$
- 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:
Have no bends at the whole real axis