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A critical point of a function occurs where the function's derivative is zero or undefined. These points are crucial because they can indicate potential maxima or minima. To find the critical points for the function \(g(s) = \frac{s^2}{s+1}\), we took its derivative using the quotient rule. The derivative is \(g'(s) = \frac{s^2 + 2s}{(s+1)^2}\). Setting this derivative equal to zero helps us identify where the function's slope is zero, which is indicative of a horizontal tangent line.
Solving \(s^2 + 2s = 0\) by factoring gives us critical points at \(s = 0\) and \(s = -2\). However, since we are considering the interval \(-\frac{1}{2} \leq s \leq 1\), only the critical point \(s = 0\) lies within this range. The point \(s = -2\) is outside our interval and is therefore not considered. Identifying these points is essential for determining where a function can potentially have extreme values.

The derivative of a function provides us with valuable information about its behavior. It represents the rate of change of the function's value with respect to a change in the input. Derivatives can be calculated using various rules, one of the most common being the quotient rule for functions in the form \(\frac{f(x)}{g(x)}\).
For our function, \(g(s) = \frac{s^2}{s+1}\), applying the quotient rule, we get: \(g'(s) = \frac{(2s)(s+1) - (s^2)(1)}{(s+1)^2} = \frac{2s^2 + 2s - s^2}{(s+1)^2} = \frac{s^2 + 2s}{(s+1)^2}\). This derivative helps us find where the slope of the curve is zero (critical points), indicating possible maxima, minima, or points of inflection. By focusing on the derivative, students better understand how the function behaves and where to look for potential extreme values.

Endpoints are the values of the function at the boundaries of the interval under consideration. For the function \(g(s) = \frac{s^2}{s+1}\) on the interval \(-\frac{1}{2} \leq s \leq 1\), the endpoints are \(s = -\frac{1}{2}\) and \(s = 1\).
Evaluating the function at these endpoints helps us determine the function's values at these critical boundary points. This is important because absolute maxima and minima can occur at these endpoints as well as at critical points.
In our case, calculating \(g(-\frac{1}{2})\) gives us \(g\left(-\frac{1}{2}\right) = \frac{\left(-\frac{1}{2}\right)^2}{-\frac{1}{2}+1} = \frac{1/4}{1/2} = \frac{1}{2}\). Similarly, for \(s = 1\), we have \(g(1) = \frac{1^2}{1+1} = \frac{1}{2}\). These evaluations confirm how the function behaves at the interval's boundaries, providing essential information for determining the absolute maximum and minimum values.

Evaluating a function at specific points, such as critical points and endpoints, is a pivotal step in finding the absolute maximum and minimum values. For our function \(g(s) = \frac{s^2}{s+1}\), we evaluated it at the critical point \(s = 0\) and the endpoints \(s = -\frac{1}{2}\) and \(s = 1\).
The evaluation process is straightforward: substitute the specific values back into the function and compute the result. For \(s = 0\), we get \(g(0) = \frac{0^2}{0+1} = 0\). At the endpoints, we have already computed \(g(-\frac{1}{2}) = \frac{1}{2}\) and \(g(1) = \frac{1}{2}\).
By comparing these values, we determine that the absolute maximum value of the function on the interval is \( \frac{1}{2}\), and it occurs both at \(s = -\frac{1}{2}\) and \(s = 1\). The absolute minimum value is \(0\) at \(s = 0\). Evaluation thus combines all previous steps to pinpoint the exact extreme values of the function.