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$$\frac{- \frac{x \left(x - 41\right) \left(\frac{\tan^{2}{\left(\frac{x}{4} - 1 \right)} + 1}{\tan^{2}{\left(\frac{x}{4} - 1 \right)}} - 1\right) \left(\tan^{2}{\left(\frac{x}{4} - 1 \right)} + 1\right)}{4} + \frac{\left(2 x - 41\right) \left(\tan^{2}{\left(\frac{x}{4} - 1 \right)} + 1\right)}{\tan{\left(\frac{x}{4} - 1 \right)}} - 4}{\tan{\left(\frac{x}{4} - 1 \right)}} = 0$$
Solve this equationThe roots of this equation
$$x_{1} = -26.5963963807254$$
$$x_{2} = -77.3407481441777$$
$$x_{3} = -13.4540762205918$$
$$x_{4} = 85.1366494898627$$
$$x_{5} = -89.9486786143009$$
$$x_{6} = 97.8056004843674$$
$$x_{7} = 8.9821143195463$$
$$x_{8} = 72.3971878134068$$
$$x_{9} = -39.3844701884795$$
$$x_{10} = -52.073306226812$$
$$x_{11} = -64.7186925945443$$
$$x_{12} = 44.516168977407$$
$$x_{13} = 59.4558084008482$$
$$x_{14} = -102.547348755834$$
$$x_{15} = 23.0549173961501$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 4$$
$$\lim_{x \to 4^-}\left(\frac{- \frac{x \left(x - 41\right) \left(\frac{\tan^{2}{\left(\frac{x}{4} - 1 \right)} + 1}{\tan^{2}{\left(\frac{x}{4} - 1 \right)}} - 1\right) \left(\tan^{2}{\left(\frac{x}{4} - 1 \right)} + 1\right)}{4} + \frac{\left(2 x - 41\right) \left(\tan^{2}{\left(\frac{x}{4} - 1 \right)} + 1\right)}{\tan{\left(\frac{x}{4} - 1 \right)}} - 4}{\tan{\left(\frac{x}{4} - 1 \right)}}\right) = -\infty$$
$$\lim_{x \to 4^+}\left(\frac{- \frac{x \left(x - 41\right) \left(\frac{\tan^{2}{\left(\frac{x}{4} - 1 \right)} + 1}{\tan^{2}{\left(\frac{x}{4} - 1 \right)}} - 1\right) \left(\tan^{2}{\left(\frac{x}{4} - 1 \right)} + 1\right)}{4} + \frac{\left(2 x - 41\right) \left(\tan^{2}{\left(\frac{x}{4} - 1 \right)} + 1\right)}{\tan{\left(\frac{x}{4} - 1 \right)}} - 4}{\tan{\left(\frac{x}{4} - 1 \right)}}\right) = \infty$$
- the limits are not equal, so
$$x_{1} = 4$$
- 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
$$\left[97.8056004843674, \infty\right)$$
Convex at the intervals
$$\left(-\infty, -102.547348755834\right]$$