91Ó°ÊÓ

The first derivative of a function helps us understand how the function behaves. It tells us the rate at which the function's value is changing. For instance, if we have a function \(f(x) = 5 - x - x^2\), the first derivative, denoted as \(f'(x)\), is found by differentiating the function with respect to \(x\).

This differentiation process gives us:
\[ f'(x) = -1 - 2x \]

By examining the first derivative, we can locate critical points, which are essential for identifying relative extrema (maximums and minimums). These points occur where \(f'(x) = 0\) or where \(f'(x)\) does not exist (though, in this case, it's always defined).

To find these points, we solve:
\[ -1 - 2x = 0 \] which simplifies to \[ x = -\frac{1}{2} \]
This critical point \( x = -\frac{1}{2} \) is essential in determining the behavior of our function.

Once we have our critical points from the first derivative, we use the second derivative to classify them as relative maxima, minima, or saddle points. The second derivative of a function gives us information about the concavity of the function.

To continue with our example, we find the second derivative of \(f(x) = 5 - x - x^2\):
\[ f''(x) = -2 \]

In this case, the second derivative is constant and negative \( f''(x) = -2 \). This tells us that the function is concave downwards everywhere.
This concavity indicates that any critical point we found will be a relative maximum.

The fact that \( f''(x)\) is less than zero at the critical point \( x = -\frac{1}{2} \), confirms that it is indeed a relative maximum.

Critical points are values of \(x\) where the first derivative of a function is zero or undefined. These points are crucial in the study of calculus because they help identify places where the function's slope changes direction.

To find critical points, we set the first derivative equal to zero and solve for \(x\). For the function \(f(x) = 5 - x - x^2 \), we set \( f'(x) = -1 - 2x = 0 \), solving we get:
\[ x = -\frac{1}{2} \]

The critical point is found at \( x = -\frac{1}{2} \). To determine the nature of this point, we use the second derivative, which tells us if it is a maximum or minimum. In this exercise, the second derivative is always negative, indicating a maximum.

After finding the critical points and testing them, we know where and what type of relative extrema the function has.

A relative maximum is a point where the function reaches a high point in its immediate vicinity, meaning higher than any other nearby points.

In our given function \( f(x) = 5 - x - x^2 \), we found the point \( x = -\frac{1}{2}\) to be critical. Using the second derivative test, we confirmed that this point is indeed a relative maximum.

Now, we need to determine the function's value at this point to complete our analysis:
\[ f \left(-\frac{1}{2} \right) = 5 - \left(-\frac{1}{2} \right) - \left(\frac{-1}{2}\right)^2 \ = 5 + \frac{1}{2} - \frac{1}{4} \ = \frac{20}{4} + \frac{2}{4} - \frac{1}{4} \ = \frac{21}{4} \] Thus, the relative maximum is at \( x = -\frac{1}{2}\) with a maximum value of \( \frac{21}{4} \).

This information is useful for graphing the function and understanding its behavior around this high point.