TeX Quellcode:

\begin{array}{rclll} f''(x) &=& \left(\left(512x^6-576x^4+162x^2\right)\ln(10)-48x^2+9\right)\cdot 10^{-8x^4+9x^2} \cr\cr f''\left(-\dfrac{3}{4}\right) &=& \left(\left(512\cdot\left(-\dfrac{3}{4}\right)^6-576\cdot\left(-\dfrac{3}{4}\right)^4+162\cdot\left(-\dfrac{3}{4}\right)^2\right)\ln(10)-48\cdot\left(-\dfrac{3}{4}\right)^2+9\right)\cdot 10^{-8\cdot\left(-\frac{3}{4}\right)^4+9\cdot\left(-\frac{3}{4}\right)^2} \quad \approx \quad -6.116{,}77 \quad < \quad 0 & & \Rightarrow \text{Hochpunkt} \cr\cr f''(0) &=& \left(\left(512\cdot 0^6-576\cdot 0^4+162\cdot 0^2\right)\ln(10)-48\cdot 0^2+9\right)\cdot 10^{-8\cdot 0^4+9\cdot 0^2} \quad = \quad 9 \quad > \quad 0 & & \Rightarrow \text{Tiefpunkt} \cr \cr f''\left(\dfrac{3}{4}\right) &=& \left(\left(512\cdot\left(\dfrac{3}{4}\right)^6-576\cdot\left(\dfrac{3}{4}\right)^4+162\cdot\left(\dfrac{3}{4}\right)^2\right)\ln(10)-48\cdot\left(\dfrac{3}{4}\right)^2+9\right)\cdot 10^{-8\cdot\left(\frac{3}{4}\right)^4+9\cdot\left(\frac{3}{4}\right)^2} \quad \approx \quad -6.116{,}77 \quad < \quad 0 & & \Rightarrow \text{Hochpunkt} \end{array}