13. Find the maximum and minimum values of the function $f(x)=(x+1)(x+2)(x+3)(x+4)+24$ on the closed interval $[-2,2]$.


$\begin{array}{l}\text { 13. } f(x)=[(x+1)(x+4)][(x+2)(x+3)]+24=\left(x^{2}+5 x+4\right)\left(x^{2}+5 x+6\right)+24=\left(x^{2}+5 x\right)^{2} \\ +10\left(x^{2}+5 x\right)+48. \\ \text { Let } t=x^{2}+5 x, x \in[-2,2]. \text { Then } t \in[-6,14], f(t)=t^{2}+10 t+48=(t+5)^{2}+23, f(t)_{\text {min }}= \\ f(-5)=23, f(t)_{\text {max }}=f(14)=384.\end{array}$