1. **Problem Statement:** Find the extreme points (maxima and minima) of a function. 2. **Formula and Rules:** To find extreme points of a function $f(x)$, we use the first derivative test. - Find the first derivative $f'(x)$. - Solve $f'(x) = 0$ to find critical points. - Use the second derivative $f''(x)$ to determine the nature of each critical point: - If $f''(x) > 0$, the point is a local minimum. - If $f''(x) < 0$, the point is a local maximum. - If $f''(x) = 0$, the test is inconclusive. 3. **Intermediate Work:** - Example: Suppose $f(x) = x^3 - 3x^2 + 4$. - Compute $f'(x) = 3x^2 - 6x$. - Set $f'(x) = 0$: $3x^2 - 6x = 0 \Rightarrow 3x(x - 2) = 0$. - Critical points are $x = 0$ and $x = 2$. - Compute $f''(x) = 6x - 6$. - Evaluate $f''(0) = 6(0) - 6 = -6 < 0$ so $x=0$ is a local maximum. - Evaluate $f''(2) = 6(2) - 6 = 6 > 0$ so $x=2$ is a local minimum. 4. **Final Answer:** The function has a local maximum at $x=0$ and a local minimum at $x=2$. This method applies generally to find extreme points of differentiable functions.