Problem 28 - University Calculus: Early Transcendentals

Question

a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying where they occur. $$g(x)=x^{4}-4 x^{3}+4 x^{2}$$

Step-by-Step Solution

Answer

Increasing on \((-\infty, 0), (1, 2), (2, \infty)\); decreasing on \((0, 1)\). Local max at \(x = 0\), local min at \(x = 1\).

Step 1: Find the First Derivative

To determine where the function is increasing or decreasing, we first find the derivative of \( g(x) = x^4 - 4x^3 + 4x^2 \). The derivative \( g'(x) \) represents the rate of change of the function:\[g'(x) = \frac{d}{dx}(x^4 - 4x^3 + 4x^2) = 4x^3 - 12x^2 + 8x.\]

Step 2: Factor the First Derivative

Next, we factor the first derivative to help find critical points:\[g'(x) = 4x(x^2 - 3x + 2).\]Further factoring gives:\[g'(x) = 4x(x-1)(x-2).\]

Step 3: Find Critical Points

Set \( g'(x) = 0 \) and solve for \( x \) to find the critical points:\[4x(x-1)(x-2) = 0.\]This yields solutions: \(x = 0, x = 1, \text{ and } x = 2.\)

Step 4: Analyze Intervals Using Critical Points

We use the critical points to test intervals and determine where \( g(x) \) is increasing or decreasing. Choose test points in the intervals defined by the critical points:- For \((-\infty, 0)\), test with \(x = -1\), \(g'(-1) > 0\).- For \((0, 1)\), test with \(x = 0.5\), \(g'(0.5) < 0\).- For \((1, 2)\), test with \(x = 1.5\), \(g'(1.5) > 0\).- For \((2, \infty)\), test with \(x = 3\), \(g'(3) > 0\).Thus, \( g(x) \) is increasing on \((-\infty, 0)\), \((1, 2)\), and \((2, \infty)\), and decreasing on \((0, 1)\).

Step 5: Determine Local and Absolute Extrema

Determine the nature of critical points using the First Derivative Test:- At \(x = 0\), the function changes from increasing to decreasing, indicating a local maximum.- At \(x = 1\), the function changes from decreasing to increasing, indicating a local minimum.- At \(x = 2\), the function continues increasing, confirming no local extrema.Evaluate \( g(x) \) at critical points:\[g(0) = 0, \quad g(1) = 1^4 - 4 \cdot 1^3 + 4 \cdot 1^2 = 1, \quad\text{and} \quad g(2) = 0.\]Thus, the local maximum is at \( x = 0 \) with value 0, and the local minimum is at \( x = 1 \) with value 1.

Key Concepts

Critical Points

Critical points are fundamental in calculus for understanding where and how a function changes. They occur where the derivative of the function, in this case, the derivative of \( g(x) = x^4 - 4x^3 + 4x^2 \), equals zero or is undefined. Identifying these points helps us analyze the behavior of the function in terms of where it increases or decreases. For the function given, the derivative is \( g'(x) = 4x(x-1)(x-2) \).
To find critical points, set \( g'(x) = 0 \):

• Factor the derivative: \( 4x(x-1)(x-2) \).
• Find the solutions to this equation: \( x = 0, x = 1, x = 2 \)

These solutions are the critical points of the function, where changes in increasing or decreasing behavior may occur.

First Derivative Test

The First Derivative Test is a method used to classify critical points and determine intervals of increase and decrease. It involves calculating the derivative, checking its sign changes around the critical points, and determining the nature of these turning points. After identifying critical points from \( g'(x) \), the next step is to choose a number in each of the intervals formed by these points:

• For the interval \((-\infty, 0)\), choosing \(x = -1\) gives \( g'(-1) > 0 \). So the function is increasing here.
• In \((0, 1)\), \(x = 0.5\) results in \( g'(0.5) < 0 \), which means the function is decreasing.
• For \((1, 2)\), \(x = 1.5\) and \( g'(1.5) > 0 \). The function is increasing again.
• Lastly, in \((2, \infty)\), \(x = 3\) gives \( g'(3) > 0 \), indicating the function is still increasing.

By examining the sign of the derivative on these intervals, the test confirms whether each critical point is a local maximum, minimum, or neither. At \(x = 0\), the function turns from increasing to decreasing, indicating a local maximum. At \(x = 1\), the function turns from decreasing to increasing, indicating a local minimum.

Local Extrema

Local extrema refer to the local maximum and minimum values within a certain range around a critical point. These are essential for understanding the peaks and valleys of the function. At these values, the function reaches a highest or lowest point relative to its immediate surroundings. For \( g(x) \):

• The local maximum is at \( x = 0 \) where the function transitions from an increasing to a decreasing trend, leading to \( g(0) = 0 \).
• There's a local minimum at \( x = 1 \), where it changes from decreasing to increasing, giving us \( g(1) = 1 \).
• At \( x = 2 \), no local extrema occur because the function continues to increase.

Evaluating these points, we understand the behavior of the function in these critical areas. They highlight where the function achieves certain limits within its open intervals.

Increasing and Decreasing Intervals

Understanding where a function is increasing or decreasing gives insights into its overall shape and trends over specific domains. These intervals are determined using the sign of the first derivative. After factoring and solving \( g'(x) \), we found critical points. By testing points within the resulting intervals, we can identify:

• The function \( g(x) \) is increasing on \( (-\infty, 0) \), \( (1, 2) \), and \( (2, \infty) \).
• It is decreasing on \( (0, 1) \).

These intervals highlight exactly where the function's rate of change is positive or negative. This is crucial for graphing and understanding the function's overall behavior. It's also instrumental in identifying where turning points such as local maxima and minima occur.