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$$- \frac{4 \cdot \left(2 x + \left(1 - \frac{4 \left(x - 1\right)^{2}}{x \left(x - 2\right)}\right) \left(x + 1\right) - 2\right)}{x^{2} \left(x - 2\right)^{2}} = 0$$
Solve this equationThe roots of this equation
$$x_{1} = - 3^{\frac{2}{3}} - \sqrt[3]{3} - 1$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0$$
$$x_{2} = 2$$
$$\lim_{x \to 0^-}\left(- \frac{4 \cdot \left(2 x + \left(1 - \frac{4 \left(x - 1\right)^{2}}{x \left(x - 2\right)}\right) \left(x + 1\right) - 2\right)}{x^{2} \left(x - 2\right)^{2}}\right) = \infty$$
Let's take the limit$$\lim_{x \to 0^+}\left(- \frac{4 \cdot \left(2 x + \left(1 - \frac{4 \left(x - 1\right)^{2}}{x \left(x - 2\right)}\right) \left(x + 1\right) - 2\right)}{x^{2} \left(x - 2\right)^{2}}\right) = -\infty$$
Let's take the limit- the limits are not equal, so
$$x_{1} = 0$$
- is an inflection point
$$\lim_{x \to 2^-}\left(- \frac{4 \cdot \left(2 x + \left(1 - \frac{4 \left(x - 1\right)^{2}}{x \left(x - 2\right)}\right) \left(x + 1\right) - 2\right)}{x^{2} \left(x - 2\right)^{2}}\right) = -\infty$$
Let's take the limit$$\lim_{x \to 2^+}\left(- \frac{4 \cdot \left(2 x + \left(1 - \frac{4 \left(x - 1\right)^{2}}{x \left(x - 2\right)}\right) \left(x + 1\right) - 2\right)}{x^{2} \left(x - 2\right)^{2}}\right) = \infty$$
Let's take the limit- the limits are not equal, so
$$x_{2} = 2$$
- 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
$$\left[- 3^{\frac{2}{3}} - \sqrt[3]{3} - 1, \infty\right)$$
Convex at the intervals
$$\left(-\infty, - 3^{\frac{2}{3}} - \sqrt[3]{3} - 1\right]$$