TeX source:
\begin{array}{rclll} f''(x) &=& \dfrac{\left(x^2-80x+1.605\right)}{25}e^{0{,}1x^2-8x-6{,}3} \cr\cr f''(40) &=& \dfrac{\left(40^2- 80 \cdot 40+1.605\right)}{25}e^{0{,}1 \cdot 40^2-8 \cdot 40-6{,}3} \quad = \quad \dfrac{1}{5} e^{-166{,}3} \quad > \quad 0 & & \Rightarrow \text{Tiefpunkt}\end{array}