91Ó°ÊÓ

To find the absolute maximum and minimum values of a function, we first need to understand the concept of a derivative. The derivative of a function gives us the slope of the function at any given point. In simple terms, it helps us understand how the function is changing at different points.
When we have a function, say \[ f(x)=\frac{1}{3} x^3-\frac{1}{2} x^2-2 x+1 \], finding the derivative involves applying standard differentiation rules to each term in the function.
For example, the derivative of \[ \frac{d}{dx} \left ( \frac{1}{3} x^3-\frac{1}{2} x^2-2 x+1 \right )\] is simply \[ f'(x) = x^2 - x - 2 \]. This helps in finding the slope of the function, which is crucial in determining the critical points.

Critical points of a function are the points where the derivative is zero or undefined. These points are important because they can indicate either maximums, minimums, or inflection points of the function.
To find the critical points of the function \[ f(x)=\frac{1}{3} x^3-\frac{1}{2} x^2-2 x+1 \], we set its derivative to zero: \[ f'(x) = x^2 - x - 2 = 0 \].
To solve this quadratic equation, we factorize it: \[ (x - 2)(x + 1) = 0 \].
Thus, the critical points are \[ x = 2 \] and \[ x = -1 \]. These points are where the function potentially has maximum or minimum values.

To determine if the critical points are maximum or minimum values, we must evaluate the original function at these points. Let's evaluate it at \[ x = -1 \] and \[ x = 2 \].
For \[ x = -1 \]:
\[ f(-1) = \frac{1}{3} (-1)^3 - \frac{1}{2} (-1)^2 - 2(-1) + 1 = \frac{-1}{3} - \frac{1}{2} + 2 + 1 = \frac{5}{6} \].
For \[ x = 2 \]:
\[ f(2) = \frac{1}{3} (2)^3 - \frac{1}{2} (2)^2 - 2(2) + 1 = \frac{8}{3} - 2 - 4 + 1 = -\frac{1}{3} \].
By comparing these values, we find that \[ f(-1) = \frac{5}{6} \] is the maximum and \[ f(2) = -\frac{1}{3} \] is the minimum.

The behavior of a function at infinity helps us understand how the function behaves as \[ x \] approaches very large positive or negative values. For the function \[ f(x)=\frac{1}{3} x^3-\frac{1}{2} x^2-2 x+1 \], the cubic term \[ \frac{1}{3}x^3 \] dominates.
As \[ x \to \infty \], \[ \frac{1}{3}x^3 \] also approaches infinity, pulling the overall function value towards infinity.
Conversely, as \[ x \to -\infty \], the \[ \frac{1}{3} x^3 \] term makes the function approach negative infinity. This tells us that there are no absolute extrema at the endpoints of infinity.
Therefore, the behavior at infinity helps confirm that the absolute maximum and minimum values occur at the critical points we found earlier.