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Piecewise functions are mathematical expressions defined by multiple sub-functions, each with its own domain. This type of function allows us to describe a system that behaves differently across different intervals of the independent variable. In our exercise, the piecewise function is defined by two expressions:

• For \( x \leq 1 \), it follows the equation \(-x^2 - 2x + 4\).
• For \( x > 1 \), it follows \( -x^2 + 6x - 4 \).

When working with piecewise functions, it's important to capture the correct conditions for each segment to find the precise domain and critical points. These critical points emerge where the behavior of the function might change, such as at the boundaries or intersection of the segments. In our example, observe that the domain of the function covers all real numbers (-∞, ∞), but the function's behavior changes at \( x = 1 \), which is a critical boundary to consider.

Derivatives are crucial for finding critical points in any function, including piecewise ones. The derivative represents the rate of change of a function and is a foundational tool in calculus. For our piecewise function, calculating the derivative for each section helps identify points where the slope is zero, indicating potential maximum or minimum values.

• For \( x \leq 1 \), the derivative \( -2x - 2 \) is computed from the sub-function \( -x^2 - 2x + 4 \). Setting the derivative equal to zero, we find a critical point at \( x = -1 \).
• For \( x > 1 \), the derivative \( -2x + 6 \) comes from \( -x^2 + 6x - 4 \), which becomes zero at \( x = 3 \).

By solving \( -2x - 2 = 0 \) and \( -2x + 6 = 0 \), critical points are identified where the function's behavior potentially shifts from increasing to decreasing or vice versa. Evaluating the original function at these points and alongside the domain endpoint provides insights into the function's graph.

Extreme values in a function are either absolute or local, which denote the highest or lowest points the function reaches within its specified domain or on a small interval. In our exercise, once we've found the critical points, we evaluatethe function's value at these places and the domain endpoint to demonstrate:

• At \( x = -1 \) and \( x = 3 \), the function value is 5. This means those points are potential candidates for absolute maximums.
• At \( x = 1 \), the function value is 1, indicating a potential candidate for an absolute minimum.

Thus, the absolute maximum is indeed 5 at \( x = -1 \) and \( x = 3 \). The absolute minimum is confirmed as 1 at \( x = 1 \). Since no other lower points are found, there is no local minimum in the typical sense of a single peak or trough within a neighborhood.