1. The problem is to find the Extrempunkte (extreme points) of a function, which are points where the function reaches a local maximum or minimum. 2. To find these points, we use the first derivative test. The formula is to find where the first derivative $f'(x)$ equals zero: $$f'(x) = 0$$ 3. Important rules: - At an Extrempunkt, the slope of the tangent (the derivative) is zero. - After finding critical points where $f'(x) = 0$, use the second derivative $f''(x)$ to determine if it is a maximum ($f''(x) < 0$) or minimum ($f''(x) > 0$). 4. Steps to find Extrempunkte: - Differentiate the function to get $f'(x)$. - Solve $f'(x) = 0$ to find critical points. - Calculate $f''(x)$ and evaluate it at each critical point. - If $f''(x) > 0$, the point is a local minimum; if $f''(x) < 0$, it is a local maximum. 5. Example: For $f(x) = x^3 - 3x^2 + 4$ - $f'(x) = 3x^2 - 6x$ - Solve $3x^2 - 6x = 0$ which simplifies to $3x\cancel{x - 2} = 0$ giving $x=0$ or $x=2$ - $f''(x) = 6x - 6$ - Evaluate $f''(0) = -6 < 0$ so $x=0$ is a local maximum - Evaluate $f''(2) = 6(2) - 6 = 6 > 0$ so $x=2$ is a local minimum 6. Thus, the Extrempunkte are at $x=0$ (maximum) and $x=2$ (minimum).