Problem 10 - Calculus Single Variable

Question

Use the first derivative to determine the intervals on which the given function \(f\) is increasing and on which \(f\) is decreasing. At each point \(c\) with \(f^{\prime}(c)=0,\) use the First Derivative Test to determine whether \(f(c)\) is a local maximum value, a local minimum value, or neither. $$ f(x)=x(x+2)^{2} $$

Step-by-Step Solution

Answer

The function is increasing on \((-\infty, -2)\) and \((-2/3, \infty)\), decreasing on \((-2, -2/3)\), with a local maximum at \(x = -2\) and a local minimum at \(x = -2/3\).

Step 1: Find the First Derivative

First, we need to differentiate the function \( f(x) = x(x+2)^2 \) to find \( f'(x) \). Using the product rule, \( (u imes v)' = u' imes v + u imes v' \), where \( u = x \) and \( v = (x+2)^2 \). First, find \( u' = 1 \) and \( v' \) using the chain rule as \( 2(x+2) \). Thus, \( f'(x) = 1 \cdot (x+2)^2 + x \cdot 2(x+2) \). Simplifying gives \( f'(x) = (x+2)^2 + 2x(x+2) \). On expanding and simplifying the terms, you get \( f'(x) = (x+2)^2 + 2x^2 + 4x = 3x^2 + 8x + 4 \).

Step 2: Find Critical Points

Set \( f'(x) = 0 \) to find the critical points: \( 3x^2 + 8x + 4 = 0 \). Solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = 8 \), and \( c = 4 \). Calculating the discriminant, you get \( b^2 - 4ac = 64 - 48 = 16 \). Thus, \( x = \frac{-8 \pm \sqrt{16}}{6} \), which simplifies to \( x= \frac{-8 \pm 4}{6} \). Hence, the critical points are \( x=-2/3 \) and \( x=-2 \).

Step 3: Analyze Intervals with Test Points

Choose test points from the intervals determined by the critical points \( x = -2 \) and \( x = -2/3 \). These intervals are \((-\infty, -2)\), \((-2, -2/3)\), and \((-2/3, \infty)\). Use test points like \( x = -3 \) for the first interval, \( x = -1 \) for the second, and \( x = 0 \) for the third. Substitute these values into \( f'(x) \). For \( x = -3 \), \( f'(-3) = 3(-3)^2 + 8(-3) + 4 = 27 - 24 + 4 = 7 \), which is positive, indicating the function is increasing. For \( x = -1 \), \( f'(-1) = 3(-1)^2 + 8(-1) + 4 = 3 - 8 + 4 = -1 \), indicating the function is decreasing. For \( x = 0 \), \( f'(0) = 3(0)^2 + 8(0) + 4 = 4 \), indicating the function is increasing.

Step 4: Classify the Critical Points

Use the First Derivative Test to classify the critical points. For \( x = -2 \), \( f'(x) \) changes from positive to negative, indicating a local maximum at \( x = -2 \). For \( x = -2/3 \), \( f'(x) \) changes from negative to positive, indicating a local minimum at \( x = -2/3 \).

Key Concepts

Derivative

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.

Increasing and Decreasing Intervals

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

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.

Quadratic Formula

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.