Para hallar los extremos hay que resolver la ecuación
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(la derivada es igual a cero),
y las raíces de esta ecuación serán los extremos de esta función:
$$\frac{d}{d x} f{\left(x \right)} = $$
primera derivada$$\frac{\frac{\left(x + 1\right) \left(\left(2 \operatorname{re}{\left(\frac{\Gamma\left(2 x + 3\right) \operatorname{polygamma}{\left(0,2 x + 3 \right)}}{\left(2 x\right)!}\right)} - 2 \operatorname{re}{\left(\frac{\left(2 x + 2\right)! \Gamma\left(2 x + 1\right) \operatorname{polygamma}{\left(0,2 x + 1 \right)}}{\left(2 x\right)!^{2}}\right)}\right) \operatorname{re}{\left(\frac{\left(2 x + 2\right)!}{\left(2 x\right)!}\right)} + \left(2 \operatorname{im}{\left(\frac{\Gamma\left(2 x + 3\right) \operatorname{polygamma}{\left(0,2 x + 3 \right)}}{\left(2 x\right)!}\right)} - 2 \operatorname{im}{\left(\frac{\left(2 x + 2\right)! \Gamma\left(2 x + 1\right) \operatorname{polygamma}{\left(0,2 x + 1 \right)}}{\left(2 x\right)!^{2}}\right)}\right) \operatorname{im}{\left(\frac{\left(2 x + 2\right)!}{\left(2 x\right)!}\right)}\right) \left(2 x\right)! \operatorname{sign}{\left(\frac{\left(2 x + 2\right)!}{\left(2 x\right)!} \right)}}{\left(2 x + 2\right)!} + \left|{\frac{\left(2 x + 2\right)!}{\left(2 x\right)!}}\right|}{x} - \frac{\left(x + 1\right) \left|{\frac{\left(2 x + 2\right)!}{\left(2 x\right)!}}\right|}{x^{2}} = 0$$
Resolvermos esta ecuaciónSoluciones no halladas,
tal vez la función no tenga extremos