91Ó°ÊÓ

Critical points are the values of \(x\) where the derivative of a function is zero or undefined. These points are crucial in determining local maximums, minimums, or points of inflection. To find critical points for a piecewise function, you must examine each piece separately.

1. **For the piece defined by** \( f(x) = 4x - 2 \): Take the derivative \( f'(x) = 4 \), which is a constant. Since it's non-zero, this segment does not have any critical points.
2. **For the piece defined by** \( f(x) = (x-2)(x-3) \): After expanding, it becomes \( x^2 - 5x + 6 \). The derivative is \( f'(x) = 2x - 5 \). Setting \( f'(x) = 0 \) gives \( x = \frac{5}{2} \).

This is a critical point. Always check if critical points lie within the given interval, \( \left[ \frac{1}{2}, \frac{7}{2} \right] \). Here, \( x = \frac{5}{2} \) does fall into \( \left[1, \frac{7}{2}\right] \), so it's a valid critical point to consider.

The absolute maximum is the greatest value that a function attains within a given interval, while the absolute minimum is the smallest value. For a piecewise function, consider the values of the function at endpoints, critical points, and potential discontinuities.

In this exercise, we look at:

• **Endpoints:** Evaluate the function at \(x = \frac{1}{2}\) and \(x = \frac{7}{2}\).
• **Critical point:** \(x = \frac{5}{2}\).
• **Point of discontinuity:** Where the function's definition changes, \(x = 1\).

After evaluating:
- \(f\left(\frac{1}{2}\right) = 0\), \(f(1) = 2\), \(f\left(\frac{5}{2}\right) = -\frac{1}{4}\), \(f\left(\frac{7}{2}\right) = \frac{3}{4}\)
The absolute maximum is \(f(1) = 2\) and the absolute minimum is \(f\left(\frac{5}{2}\right) = -\frac{1}{4}\).

Evaluating derivatives is essential for identifying critical points and understanding the behavior of a function. Derivatives represent the rate of change, or slope, of a function at a given point.

To evaluate a derivative:

• **For linear functions like** \(f(x) = 4x - 2\), the derivative is simply the coefficient of \(x\), which is 4.
• **For polynomial functions like** \(f(x) = (x-2)(x-3)\): Expand to \(f(x) = x^2 - 5x + 6\) before finding the derivative \(f'(x) = 2x - 5\).

Set the derivative equal to zero to find critical points. There is a critical point at \(x = \frac{5}{2}\) since \(f'(x) = 0\) there.

Interval notation is a way of representing a range of values that \(x\) can take. It includes endpoints and signifies whether they are part of the interval or not.

1. **Closed interval:** \([a, b]\) means \(x\) can be \(a\), \(b\), or any value in between. The square brackets indicate that the endpoints are included.
2. **Open interval:** \((a, b)\) indicates \(x\) can be any value between \(a\) and \(b\), but not \(a\) or \(b\) themselves. Curved brackets are used to exclude endpoints.

In this exercise, the interval \(\left[ \frac{1}{2}, \frac{7}{2} \right]\) means we're considering all values from \(\frac{1}{2}\) to \(\frac{7}{2}\) inclusively. It's vital to only assess points from within this range, including endpoints, when determining absolute maximum and minimum values.