Problem 28 - Thomas Calculus

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 \((0, 1)\) and \((2, fty)\); decreasing on \((- fty, 0)\) and \((1, 2)\). Local min at \(x=0\) and \(x=2\) (value 0); local max at \(x=1\) (value 1).

Step 1: Find the First Derivative

To determine the intervals on which the function is increasing or decreasing, first find the first derivative of \( g(x) \). Given \( g(x) = x^4 - 4x^3 + 4x^2 \), find \( g'(x) \) by using basic differentiation rules: \( g'(x) = \frac{d}{dx}(x^4) - \frac{d}{dx}(4x^3) + \frac{d}{dx}(4x^2) \). This gives \( g'(x) = 4x^3 - 12x^2 + 8x \).

Step 2: Set the First Derivative to Zero

To find critical points, set the first derivative \( g'(x) = 4x^3 - 12x^2 + 8x \) equal to zero: \( 4x^3 - 12x^2 + 8x = 0 \). Factor out the greatest common factor: \( 4x(x^2 - 3x + 2) = 0 \). Further factorize to get: \( 4x(x-1)(x-2) = 0 \). This gives critical points at \( x = 0, 1, 2 \).

Step 3: Test Intervals for Increase and Decrease

Choose test points in each interval determined by the critical points to determine if the function \( g(x) \) is increasing or decreasing. Test intervals are \((-fty, 0)\), \((0, 1)\), \((1, 2)\), and \((2, fty)\).- For \( x \in (-fty, 0) \), choose \( x = -1 \), then \( g'(-1) = 4(-1)^3 - 12(-1)^2 + 8(-1) = -4 - 12 - 8 = -24 \). This means \( g(x) \) is decreasing.- For \( x \in (0, 1) \), choose \( x = 0.5 \), then \( g'(0.5) = 4(0.5)^3 - 12(0.5)^2 + 8(0.5) = 1 - 3 + 4 = 2 \). This means \( g(x) \) is increasing.- For \( x \in (1, 2) \), choose \( x = 1.5 \), then \( g'(1.5) = 4(1.5)^3 - 12(1.5)^2 + 8(1.5) = 13.5 - 27 + 12 = -1.5 \). This means \( g(x) \) is decreasing.- For \( x \in (2, fty) \), choose \( x = 3 \), then \( g'(3) = 4(3)^3 - 12(3)^2 + 8(3) = 108 - 108 + 24 = 24 \). This means \( g(x) \) is increasing.

Step 4: Identify Local and Absolute Extremes

Based on the increasing and decreasing intervals, identify local minima and maxima.- At \( x=0 \), the function changes from decreasing to increasing, indicating a local minimum. Calculate \( g(0) = 0^4 - 4(0)^3 + 4(0)^2 = 0 \).- At \( x=1 \), the function changes from increasing to decreasing, indicating a local maximum. Calculate \( g(1) = 1^4 - 4(1)^3 + 4(1)^2 = 1 - 4 + 4 = 1 \).- At \( x=2 \), the function changes from decreasing to increasing, indicating another local minimum. Calculate \( g(2) = 2^4 - 4(2)^3 + 4(2)^2 = 16 - 32 + 16 = 0 \).

Step 5: Conclusion

The function \( g(x) \) is increasing on the intervals \((0, 1)\) and \((2, fty)\), and decreasing on \((-fty, 0)\) and \((1, 2)\). There is a local minimum at \( x=0 \) and \( x=2 \) with a value of 0, and a local maximum at \( x=1 \) with a value of 1.

Key Concepts

First Derivative

To find where a function is increasing or decreasing, the first step is to calculate its first derivative. The first derivative, often denoted as \( g'(x) \), provides information about the slope of the function at any given point. For the function \( g(x) = x^4 - 4x^3 + 4x^2 \), we apply differentiation rules to find \( g'(x) \). By differentiating each term individually, we have:

• \( \frac{d}{dx}(x^4) = 4x^3 \)
• \( \frac{d}{dx}(-4x^3) = -12x^2 \)
• \( \frac{d}{dx}(4x^2) = 8x \)

Therefore, the first derivative is \( g'(x) = 4x^3 - 12x^2 + 8x \). This derivative is useful because where it equals zero or is undefined, the original function might change from increasing to decreasing, or vice versa. These are called critical points.

Increasing and Decreasing Intervals

Once we have the first derivative, \( g'(x) = 4x^3 - 12x^2 + 8x \), we use it to determine the intervals where the function \( g(x) \) is increasing or decreasing. To do this, set \( g'(x) = 0 \) to find the critical points. By solving \( 4x(x-1)(x-2) = 0 \), we find critical points at \( x = 0, 1, 2 \). These points divide the x-axis into intervals, which we test:

• For \( x < 0 \), choose \( x = -1 \): \( g'(-1) = -24 \), indicating a decreasing interval.
• For \( 0 < x < 1 \), choose \( x = 0.5 \): \( g'(0.5) = 2 \), indicating an increasing interval.
• For \( 1 < x < 2 \), choose \( x = 1.5 \): \( g'(1.5) = -1.5 \), indicating a decreasing interval.
• For \( x > 2 \), choose \( x = 3 \): \( g'(3) = 24 \), indicating an increasing interval.

Therefore, \( g(x) \) is increasing on \((0, 1)\) and \((2, \infty)\), and decreasing on \((-\infty, 0)\) and \((1, 2)\).

Local Extrema

Local extrema are points where the function reaches a local maximum or minimum within an interval. By examining changes in the first derivative, we identify these points. For \( g(x) = x^4 - 4x^3 + 4x^2 \), the critical points help us:

• At \( x = 0 \), \( g'(x) \) changes from negative to positive, so it is a local minimum; \( g(0) = 0 \).
• At \( x = 1 \), \( g'(x) \) changes from positive to negative, so it is a local maximum; \( g(1) = 1 \).
• At \( x = 2 \), \( g'(x) \) changes from negative to positive again, indicating another local minimum; \( g(2) = 0 \).

Thus, the local minima occur at \( x = 0 \) and \( x = 2 \), and the local maximum occurs at \( x = 1 \).

Differentiation Rules

Differentiation rules are fundamental for finding derivatives efficiently. They simplify the process of differentiating complex functions. For example, common rules include:

• The power rule for any term \( x^n \), which is \( nx^{n-1} \).
• The constant rule, which says the derivative of a constant is 0.
• The sum rule, stating that the derivative of a sum of functions is the sum of their derivatives.

In \( g(x) = x^4 - 4x^3 + 4x^2 \), we applied these rules:

• Power rule was used for each term separately.
• By applying these rules, finding \( g'(x) \) became not only direct but also error-free.

These differentiation rules are essential for solving calculus problems and analyzing functions effectively.