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The derivative of a function is a fundamental concept in calculus that represents the rate of change of the function with respect to its variable. In simpler terms, it tells us how the value of a function changes as the input, or independent variable, changes. The derivative is often denoted as \( f'(x) \) and is determined by differentiating the original function.

To find the derivative of a function like \( f(x) = x(x+2)^2 \), we use the product rule, which is used when differentiating products of two functions. The product rule states: if you have two functions \( u \) and \( v \), the derivative of their product \( u \cdot v \) is \( u' \cdot v + u \cdot v' \).

In our case, \( u \) is \( x \), and \( v \) is \( (x+2)^2 \). Therefore, we differentiate using:

• \( u' = 1 \)
• \( v' = 2(x+2) \) using the chain rule

Combining this, the derivative \( f'(x) \) simplifies to \( 3x^2 + 8x + 4 \) after applying the product rule and simplifying the result.

Understanding where a function increases or decreases is critical when analyzing its behavior. A function is said to be increasing on an interval if the derivative, \( f'(x) \), is positive throughout that interval. Conversely, it is decreasing if \( f'(x) \) is negative.

Using the function \( f(x) = x(x+2)^2 \) and its derivative \( f'(x) = 3x^2 + 8x + 4 \), one can determine these intervals. By testing values in different intervals created around the critical points, one can see the sign (positive or negative) of the derivative:

• Pick a test point for each interval among \((-\infty, -2)\), \((-2, -2/3)\), and \((-2/3, \infty)\).
• If \( f'(x) > 0 \) at a test point, the function is increasing on that interval.
• If \( f'(x) < 0 \) at a test point, the function is decreasing on that interval.

This method clearly shows whether the function increases or decreases between its critical points, giving insights into its general shape.

Critical points in a function are values of \( x \) at which the derivative \( f'(x) \) is zero or undefined. These points are significant because they often correspond to the peaks and valleys of a function—its local maxima or minima. By locating these critical points, we can understand where the function might change direction from increasing to decreasing or vice versa.

In our example, we set the derivative \( f'(x) = 3x^2 + 8x + 4 \) to zero to find the critical points. Solving the equation gives:

• Critical Point 1: \( x = -2 \)
• Critical Point 2: \( x = -2/3 \)

The First Derivative Test is then used to classify each critical point as a local maximum, local minimum, or neither, depending on how the sign of \( f'(x) \) changes on either side of the point.

The quadratic formula is a useful tool in solving quadratic equations of the form \( ax^2 + bx + c = 0 \). This formula provides solutions for \( x \) as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

In the context of our example, the quadratic formula helps us find the critical points by solving \( 3x^2 + 8x + 4 = 0 \). Here, \( a = 3 \), \( b = 8 \), and \( c = 4 \). Calculating the discriminant \( b^2 - 4ac \) yields 16, allowing us to determine:

• \( x = -2 \) and \( x = -2/3 \)

These points are crucial for understanding the behavior of the function, as they mark where its slope is zero, leading to changes in increase and decrease behavior.