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$$\frac{2 \left(2 - \frac{3 \left(2 x + 3\right)}{x - 1}\right) \left(2 - \frac{2 x + 3}{x - 1}\right)}{\left(x - 1\right)^{2}} = 0$$
Solve this equationThe roots of this equation
$$x_{1} = - \frac{11}{4}$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 1$$
$$\lim_{x \to 1^-}\left(\frac{2 \left(2 - \frac{3 \left(2 x + 3\right)}{x - 1}\right) \left(2 - \frac{2 x + 3}{x - 1}\right)}{\left(x - 1\right)^{2}}\right) = \infty$$
$$\lim_{x \to 1^+}\left(\frac{2 \left(2 - \frac{3 \left(2 x + 3\right)}{x - 1}\right) \left(2 - \frac{2 x + 3}{x - 1}\right)}{\left(x - 1\right)^{2}}\right) = \infty$$
- 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
$$\left[- \frac{11}{4}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, - \frac{11}{4}\right]$$