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$$\frac{\sqrt{- \frac{x - 5}{x - 2}} \left(\frac{x - 5}{x - 2} - 1\right) \left(\frac{2}{x - 2} + \frac{\frac{x - 5}{x - 2} - 1}{x - 5} + \frac{2}{x - 5}\right)}{4 \left(x - 5\right)} = 0$$
Solve this equationThe roots of this equation
$$x_{1} = \frac{17}{4}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 2$$
$$\lim_{x \to 2^-}\left(\frac{\sqrt{- \frac{x - 5}{x - 2}} \left(\frac{x - 5}{x - 2} - 1\right) \left(\frac{2}{x - 2} + \frac{\frac{x - 5}{x - 2} - 1}{x - 5} + \frac{2}{x - 5}\right)}{4 \left(x - 5\right)}\right) = \infty i$$
$$\lim_{x \to 2^+}\left(\frac{\sqrt{- \frac{x - 5}{x - 2}} \left(\frac{x - 5}{x - 2} - 1\right) \left(\frac{2}{x - 2} + \frac{\frac{x - 5}{x - 2} - 1}{x - 5} + \frac{2}{x - 5}\right)}{4 \left(x - 5\right)}\right) = \infty$$
- 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
$$\left(-\infty, \frac{17}{4}\right]$$
Convex at the intervals
$$\left[\frac{17}{4}, \infty\right)$$