1. **Problem statement:** Find the local maxima and minima (Hoch- und Tiefpunkte) of the function $f(x)$ using the sign change criterion for the derivative. 2. **Given functions:** a) $f(x) = x^2 - 3x + 2$ b) $f(x) = 2x^3 - 3x^2 - 12x - 6$ c) $f(x) = x^4 - 4x + 1$ 3. **Method:** - Find the first derivative $f'(x)$. - Solve $f'(x) = 0$ to find critical points. - Use the sign change of $f'(x)$ around each critical point to determine if it is a local max, min, or neither. 4. **Calculations:** a) $f(x) = x^2 - 3x + 2$ - Derivative: $f'(x) = 2x - 3$ - Solve $2x - 3 = 0 \Rightarrow x = \frac{3}{2} = 1.5$ - Test sign of $f'(x)$ around $x=1.5$: - For $x < 1.5$, say $x=1$, $f'(1) = 2(1) - 3 = -1 < 0$ - For $x > 1.5$, say $x=2$, $f'(2) = 4 - 3 = 1 > 0$ - Since $f'(x)$ changes from negative to positive, $x=1.5$ is a local minimum. - Find $f(1.5) = (1.5)^2 - 3(1.5) + 2 = 2.25 - 4.5 + 2 = -0.25$ - **Local minimum at $(1.5, -0.25)$** b) $f(x) = 2x^3 - 3x^2 - 12x - 6$ - Derivative: $f'(x) = 6x^2 - 6x - 12$ - Solve $6x^2 - 6x - 12 = 0$: Divide by 6: $x^2 - x - 2 = 0$ Factor: $(x - 2)(x + 1) = 0$ Solutions: $x=2$ and $x=-1$ - Test sign of $f'(x)$ around $x=-1$: - For $x=-2$, $f'(-2) = 6(4) - 6(-2) - 12 = 24 + 12 - 12 = 24 > 0$ - For $x=0$, $f'(0) = 0 - 0 - 12 = -12 < 0$ - So $f'(x)$ changes from positive to negative at $x=-1$, indicating a local maximum. - Test sign of $f'(x)$ around $x=2$: - For $x=1$, $f'(1) = 6 - 6 - 12 = -12 < 0$ - For $x=3$, $f'(3) = 54 - 18 - 12 = 24 > 0$ - So $f'(x)$ changes from negative to positive at $x=2$, indicating a local minimum. - Calculate function values: - $f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) - 6 = -2 - 3 + 12 - 6 = 1$ - $f(2) = 2(8) - 3(4) - 24 - 6 = 16 - 12 - 24 - 6 = -26$ - **Local maximum at $(-1, 1)$** - **Local minimum at $(2, -26)$** c) $f(x) = x^4 - 4x + 1$ - Derivative: $f'(x) = 4x^3 - 4$ - Solve $4x^3 - 4 = 0 \Rightarrow x^3 = 1 \Rightarrow x = 1$ - Test sign of $f'(x)$ around $x=1$: - For $x=0$, $f'(0) = -4 < 0$ - For $x=2$, $f'(2) = 4(8) - 4 = 32 - 4 = 28 > 0$ - Since $f'(x)$ changes from negative to positive at $x=1$, this is a local minimum. - Calculate $f(1) = 1 - 4 + 1 = -2$ - **Local minimum at $(1, -2)$** 5. **Summary:** - a) Local minimum at $(1.5, -0.25)$ - b) Local maximum at $(-1, 1)$ and local minimum at $(2, -26)$ - c) Local minimum at $(1, -2)$