Problem 12 - 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. $$ f(x)=(x+1)^{2}(x+2)^{2} $$

Step-by-Step Solution

Answer

The function is increasing on \((-2, -\frac{3}{2})\) and \((-1, \infty)\), decreasing on \((-\infty, -2)\) and \((-\frac{3}{2}, -1)\).

Step 1: Find the First Derivative

To determine the intervals on which the function is increasing or decreasing, we need to find the first derivative of the function \( f(x) = (x+1)^2 (x+2)^2 \). Using the product rule and chain rule, find \( f'(x) \). Let \( u = (x+1)^2 \) and \( v = (x+2)^2 \), then the derivative \( f'(x) = u'v + uv' \). Calculate \( u' = 2(x+1) \) and \( v' = 2(x+2) \). Substituting these into the derivative gives: \[ f'(x) = 2(x+1)(x+2)^2 + (x+1)^2 2(x+2) \] Simplifying, we get: \[ f'(x) = 2(x+1)(x+2)[(x+2)+(x+1)] = 4(x+1)(x+2)(2x+3) \].

Step 2: Find Critical Points

The critical points occur where the first derivative \( f'(x) \) is equal to zero or undefined. Here, \( f'(x) = 0 \), so solve \[ 4(x+1)(x+2)(2x+3) = 0 \]. The solutions are given by setting each factor equal to zero: \( x+1 = 0 \), \( x+2 = 0 \), and \( 2x+3 = 0 \). Solving these equations, we find the critical points: \( x = -1, -2, -\frac{3}{2} \).

Step 3: Determine Test Intervals

Use the critical points to create intervals on the number line: \((-\infty, -2), (-2, -\frac{3}{2}), (-\frac{3}{2}, -1), (-1, \infty)\). We will test each interval to determine whether \( f(x) \) is increasing or decreasing. Select a test point from each of these intervals.

Step 4: Apply the First Derivative Test

For each interval, pick a test point, substitute it into \( f'(x) \), and determine the sign: - Interval \((-\infty, -2)\), use \( x = -3 \): \( f'(-3) = 4(-3+1)(-3+2)(2(-3)+3) = 4(-2)(-1)(-3) < 0 \), decreasing.- Interval \((-2, -\frac{3}{2})\), use \( x = -1.75 \): \( f'(-1.75) = 4(-1.75+1)(-1.75+2)(2(-1.75)+3) > 0 \), increasing.- Interval \((-\frac{3}{2}, -1)\), use \( x = -1.25 \): \( f'(-1.25) = 4(-1.25+1)(-1.25+2)(2(-1.25)+3) < 0 \), decreasing.- Interval \((-1, \infty)\), use \( x = 0 \): \( f'(0) = 4(0+1)(0+2)(2(0)+3) > 0 \), increasing.

Step 5: Determine Local Maxima and Minima

The First Derivative Test tells us:- At \( x = -2 \): Changes from decreasing to increasing, local minimum.- At \( x = -\frac{3}{2} \): Changes from increasing to decreasing, local maximum.- At \( x = -1 \): Changes from decreasing to increasing, local minimum.

Key Concepts

First Derivative Test

The First Derivative Test is a very helpful method for determining where a function has local maximum or minimum values. To use it, first find the derivative of your function. This step gives you a new function that describes the slope of the tangent lines to the original function's graph at any point. Critical points, where the derivative is zero or undefined, are crucial here. At these points, the function can change direction.

After finding the derivative, divide your x-axis into intervals based on these critical points. Choose a test point from each interval and substitute it into the derivative to determine if the function is increasing or decreasing. The First Derivative Test tells you whether a critical point is a local maximum, local minimum, or neither by observing the sign changes of the derivative around that point. If the derivative changes from positive to negative, it’s a local maximum. From negative to positive, it’s a local minimum.

Critical points

Critical points are special values of x in the domain of a function where the derivative is either zero or does not exist. These points are potential locations for local maxima or minima. Let's explore how to find these points:

• Find the first derivative of the function. This is key to determining the rate of change of the function at any x point.
• Set the derivative equal to zero and solve for x. Also, check where the derivative is undefined.
• In our exercise, solving the equation \( 4(x+1)(x+2)(2x+3) = 0 \) gives us the critical points: \( x = -1, -2, -\frac{3}{2} \).

These critical points are essential for analyzing the behavior of the function and how it changes across different intervals.

Intervals of increase and decrease

Determining where a function is increasing or decreasing enhances our understanding of its overall behavior. Once you have the first derivative and critical points:

• Create intervals on the x-axis. Use critical points to break the number line into different sections.
• Select a test point from each interval. For example, use \( x = -3 \) for the interval \((-cinfty, -2)\), and check the sign of the derivative at that point.
• If the derivative at the test point is positive, the function is increasing in that interval. If negative, the function is decreasing.

For example, in our exercise, the function is decreasing in the interval \((-cinfty, -2)\) and increasing in \((-2, -\frac{3}{2})\). This pattern repeats with increasing and decreasing intervals across the x-axis.

Local maxima and minima

Local maxima and minima are key features of any function, representing the highest or lowest values that occur in a small neighborhood of a critical point.

• Use the sign changes of the derivative around critical points. This will help identify the nature of these points.
• If the derivative changes from positive to negative at a critical point, it's a local maximum. If it changes from negative to positive, it's a local minimum.
• In our example, \(x = -2\) and \(x = -1\) are local minima because the function changes from decreasing to increasing, and \(x = -\frac{3}{2}\) is a local maximum as it changes from increasing to decreasing.

These local extrema are very telling as they can hint at the overall shape and tendencies of the function.