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$$\frac{- \frac{1}{\left(2 x - 7\right)^{\frac{3}{2}}} - \frac{4}{x \sqrt{2 x - 7}} + \frac{6 \sqrt{2 x - 7}}{x^{2}}}{x^{2}} = 0$$
Solve this equationThe roots of this equation
$$x_{1} = \frac{14}{3} - \frac{7 \sqrt{10}}{15}$$
$$x_{2} = \frac{7 \sqrt{10}}{15} + \frac{14}{3}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 0$$
$$\lim_{x \to 0^-}\left(\frac{- \frac{1}{\left(2 x - 7\right)^{\frac{3}{2}}} - \frac{4}{x \sqrt{2 x - 7}} + \frac{6 \sqrt{2 x - 7}}{x^{2}}}{x^{2}}\right) = \infty i$$
$$\lim_{x \to 0^+}\left(\frac{- \frac{1}{\left(2 x - 7\right)^{\frac{3}{2}}} - \frac{4}{x \sqrt{2 x - 7}} + \frac{6 \sqrt{2 x - 7}}{x^{2}}}{x^{2}}\right) = \infty i$$
- limits are equal, then skip the corresponding 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
$$\left[\frac{7 \sqrt{10}}{15} + \frac{14}{3}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, \frac{7 \sqrt{10}}{15} + \frac{14}{3}\right]$$