Problem 18 - Thomas Calculus

Question

In Exercises \(9-28 :\) a. Find the intervals on which the function is increasing and decreasing. b. Then identify the function's local extreme values, if any, saying where they are taken on. c. Which, if any, of the extreme values are absolute? d. Support your findings with a graphing calculator or computer grapher. $$ g(x)=x^{4}-4 x^{3}+4 x^{2} $$

Step-by-Step Solution

Answer

Increasing: (0, 2) and (2, ∞). Decreasing: (-∞, 0). Local min at x=0 and x=2, local max at x=1. Absolute min at x=0 and x=2.

Step 1: Find the Derivative

To determine where the function is increasing or decreasing, we first find the derivative of the function. We have the function \( g(x) = x^4 - 4x^3 + 4x^2 \). The derivative of \( g(x) \) is found by applying the power rule to each term: \( g'(x) = 4x^3 - 12x^2 + 8x \).

Step 2: Determine Critical Points

Critical points occur where the derivative is zero or undefined. Set the derivative equal to zero and solve: \( 4x^3 - 12x^2 + 8x = 0 \). Factor out the greatest common factor, which is \( 4x \), giving us \( 4x(x^2 - 3x + 2) = 0 \). Further factor the quadratic to get \( 4x(x-1)(x-2)=0 \). The critical points are \( x = 0, 1, 2 \).

Step 3: Test Intervals Around Critical Points

To find intervals where the function is increasing or decreasing, test intervals around the critical points. Use the first derivative test. Evaluate \( g'(x) \) over the intervals \( (-\infty, 0) \), \( (0, 1) \), \( (1, 2) \), and \( (2, \infty) \). - For \( x < 0 \), choose \( g'(-1): 4(-1)^3 - 12(-1)^2 + 8(-1) < 0 \). - For \( 0 < x < 1 \), choose \( g'(0.5): positive \). - For \( 1 < x < 2 \), choose \( g'(1.5): positive \). - For \( x > 2 \), choose \( g'(3): 4(3)^3 - 12(3)^2 + 8(3) > 0 \).

Step 4: Identify Increasing and Decreasing Intervals

From Step 3, we determine that:- The function is decreasing on \( (-\infty, 0) \).- The function is increasing on \( (0, 2) \).- The function is increasing on \( (2, \infty) \).

Step 5: Identify Local Extremes

Using the results from Step 3, evaluate the function at the critical points to determine local extremes:- At \( x=0 \), \( g(0) = 0 \). - At \( x=1 \), \( g(1) = 1 - 4 + 4 = 1 \).- At \( x=2 \), \( g(2) = 16 - 32 + 16 = 0 \).- The function has 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.

Step 6: Determine Absolute Extremes

Compare the local extremes, find that both local minima at \( x=0 \) and \( x=2 \) with a value of \( 0 \) are absolute minimum values, since they are the lowest points on the curve.- The local maximum at \( x=1 \) is not an absolute maximum since the function increases indefinitely as \( x \to \infty \).

Step 7: Verify with Graphing Tool

Use a graphing tool to confirm intervals of increase, decrease, and the locations of local extrema. The graph will show the function decreasing until \( x=0 \), increasing through \( x=1 \), decreasingly slightly, then increasing from \( x=2 \) onward. This confirms our calculated intervals and the locations of extrema.

Key Concepts

Critical Points

In calculus, critical points are where a function's derivative is zero or undefined. For function analysis, finding these points is essential because they can indicate local maxima, minima, or saddle points. In the given exercise, the function is defined as \[ g(x)=x^{4}-4x^{3}+4x^{2} \] To find the critical points, we first take the derivative: \[ g'(x) = 4x^3 - 12x^2 + 8x \] Setting the derivative equal to zero, \[ 4x^3 - 12x^2 + 8x = 0 \] we factor to get: \[ 4x(x-1)(x-2)=0 \] The solutions to this equation are the critical points of the function, specifically at \( x = 0, 1, \text{ and } 2 \). Each critical point potentially marks a significant change in the function's behavior.

Increasing and Decreasing Intervals

To understand where a function is increasing or decreasing, we consider the sign of its first derivative. When the derivative is positive, the function is increasing; when negative, the function is decreasing. We analyze the function's derivative: \[ g'(x) = 4x^3 - 12x^2 + 8x \] Using critical points \( x = 0, 1, \text{ and } 2 \), we test the sign of \( g'(x) \) in different intervals: - For \((-\infty, 0)\), \( g'(x) < 0 \), indicating the function decreases. - For \( (0, 1) \) and \( (1, 2) \), \( g'(x) > 0 \), indicating the function increases. - For \( (2, \infty) \), \( g'(x) > 0 \), meaning the function increases again. Thus, the intervals are divided into decreasing on \((-\infty, 0)\) and increasing on both \((0, 2)\) and \((2, \infty)\).

Local and Absolute Extreme Values

Local and absolute extreme values help us find points where the function reaches its peak or lowest values locally or globally. From the critical points, evaluate \( g(x) \) to find extremes: - At \( x = 0 \), \( g(0) = 0 \) is a local minimum because the derivative changes from negative to positive. - At \( x = 1 \), \( g(1) = 1 - 4 + 4 = 1 \), this is a local maximum since the derivative changes from positive to negative. - At \( x = 2 \), \( g(2) = 16 - 32 + 16 = 0 \), which is again a local minimum. Comparing these, both \( x = 0 \) and \( x = 2 \) with their function value of \( 0 \) qualify as absolute minimums because no lower values are seen in the function's curve. Meanwhile, there's no absolute maximum since the function rises indefinitely as \( x \to \infty \).

Derivative Test

The derivative test is a useful tool for identifying the behavior of a function around its critical points. We apply the first derivative test, which involves evaluating the sign changes in the first derivative \( g'(x) \). When analyzing the intervals around each critical point in the function \( g(x)=x^{4}-4x^{3}+4x^{2} \), we determine: - If \( g'(x) \) changes from positive to negative at a point, it indicates a local maximum. - If \( g'(x) \) changes from negative to positive, it showcases a local minimum. For instance, at \( x=1 \), \( g'(x) \) transitions from positive to negative, making \( x = 1 \) a local maximum. Meanwhile, at \( x=0 \) and \( x=2 \), the derivative transitions from negative to positive indicates local minima, confirming the results of our analysis.