Problem 37 - Algebra and Trigonometry

Question

\(35-42\) A function is given. (a) Find all the local maximum and minimum values of the function and the value of \(x\) at which each occurs. State each answer correct to two decimal places. (b) Find the intervals on which the function is increasing and on which the function is decreasing. State each answer correct to two decimal places. $$ g(x)=x^{4}-2 x^{3}-11 x^{2} $$

Step-by-Step Solution

Answer

Local minima at \( x = -1.68, 3.18 \); local maximum at \( x = 0 \). Increasing: \((-\infty, -1.68), (0, 3.18)\); decreasing: \((-1.68, 0), (3.18, \infty)\).

Step 1: Find the derivative

To find the local maxima or minima, we need the first derivative of the function. Let's find the derivative of \( g(x) = x^4 - 2x^3 - 11x^2 \). This involves using the power rule.\[ g'(x) = \frac{d}{dx}(x^4) - \frac{d}{dx}(2x^3) - \frac{d}{dx}(11x^2) = 4x^3 - 6x^2 - 22x \]

Step 2: Solve for critical points

Critical points occur where the first derivative is zero or undefined. We set the derivative equal to zero and solve for \( x \):\[ 4x^3 - 6x^2 - 22x = 0 \]Factor out a common term, \( x \):\[ x(4x^2 - 6x - 22) = 0 \]This gives us \( x = 0 \) and we solve the quadratic \( 4x^2 - 6x - 22 = 0 \) using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 4 \), \( b = -6 \), \( c = -22 \).

Step 3: Calculate the roots of the quadratic

Substitute the values into the quadratic formula:\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 4 \cdot (-22)}}{2 \cdot 4} \]\[ x = \frac{6 \pm \sqrt{36 + 352}}{8} \]\[ x = \frac{6 \pm \sqrt{388}}{8} \]\[ x = \frac{6 \pm 19.7}{8} \]This gives approximately \( x = 3.18 \) and \( x = -1.68 \) (rounded to two decimal places).

Step 4: Test intervals for local maxima/minima

Use these critical points \( x = 0, 3.18, -1.68 \) to determine local maxima and minima by testing intervals around them. Evaluate the sign changes of the derivative around these points to determine where the function increases or decreases.For example, test values in intervals \((-\infty, -1.68)\), \((-1.68, 0)\), \((0, 3.18)\), and \((3.18, \infty)\) to determine sign changes in \( g'(x) \).

Step 5: Determine intervals of increase and decrease

Evaluate the sign of \( g'(x) \) at test points in each interval:- For \((-\infty, -1.68)\), \( g'(x) > 0 \): increasing.- For \((-1.68, 0)\), \( g'(x) < 0 \): decreasing.- For \((0, 3.18)\), \( g'(x) > 0 \): increasing.- For \((3.18, \infty)\), \( g'(x) < 0 \): decreasing.

Step 6: Conclude local maxima and minima

Based on our test intervals, determine:- \( x = -1.68 \) and \( x = 3.18 \) as points of local minima because the derivative changes from negative to positive.- \( x = 0 \) as a point of local maximum because the derivative changes from positive to negative.Evaluate \( g(x) \) at these critical points for their respective values:- \( g(-1.68) \approx 13.45 \)- \( g(3.18) \approx -109.72 \)- \( g(0) = 0 \)

Key Concepts

Local Maximum and Minimum

Finding the local maximum and minimum of a function is a core concept in calculus, particularly useful in understanding the behavior of graphs. This is where a function peaks (maximum) or dips (minimum) in a localized region. These points help us understand where the function changes direction.

To identify these local extrema, we first need to calculate the derivative of the function. The derivative, denoted as \( g'(x) \), provides the rate of change of \( g(x) \). In our original exercise, we found the derivative to be \( g'(x) = 4x^3 - 6x^2 - 22x \).

Next, we find the critical points by setting the derivative to zero, \( g'(x) = 0 \), which reveals where the slope of the tangent to the graph is horizontal. Solving \( 4x^3 - 6x^2 - 22x = 0 \) provided us with potential candidates for local maxima or minima. These critical points we found were \( x = 0 \), \( x = -1.68 \), and \( x = 3.18 \).

After identifying the critical points, we determine which are maxima and which are minima by analyzing the sign changes of the derivative around these points. For instance, a change from positive to negative in derivative values indicates a local maximum, while a transition from negative to positive highlights a local minimum.

Increasing and Decreasing Intervals

Understanding where a function is increasing or decreasing is pivotal for sketching the graph and provides insights into the overall trend of the function.

A function is **increasing** on an interval if for any two points within this interval, a smaller input results in a smaller output. Mathematically, if \( g'(x) > 0 \), the function is rising in that interval. Conversely, if \( g'(x) < 0 \), the function is **decreasing** because a smaller input than any other in the same interval results in a larger output.

In our step-by-step solution, to determine these intervals, we evaluated \( g'(x) \) at various test points across subintervals defined by the critical points \( x = -1.68 \), \( x = 0 \), and \( x = 3.18 \). Here's what we found:

• **Increasing:** For \((-fty, -1.68)\) and \((0, 3.18)\)
• **Decreasing:** For \((-1.68, 0)\) and \((3.18, \infty)\)

Recognizing these intervals helps in comprehending the nature of turning points and ensuring any major changes in the graph's slope are noted.

First Derivative Test

The First Derivative Test is a systematic way used to classify critical points as local maximums or minimums by observing the behavior of \( g'(x) \) around these points.

Critical points arise where \( g'(x) = 0 \) or where the derivative is undefined. In this exercise, however, we only had points where it equaled zero, which were \( x = 0 \), \( x = -1.68 \), and \( x = 3.18 \).

To apply the First Derivative Test effectively, here are the steps you follow:

• Select a test point from each of the intervals created by the critical points.
• Assess the sign of \( g'(x) \) at each test location:
• If \( g'(x) \) shifts from positive to negative as you pass through a critical point, you label it a local maximum.
• If \( g'(x) \) transitions from negative to positive, it's a local minimum.

By applying this test, it's clear that \( x = 0 \) is a local maximum because \( g'(x) \) transitions from positive to negative as we pass through it. Conversely, \( x = -1.68 \) and \( x = 3.18 \) are local minima, characterized by a switch from negative to positive in \( g'(x) \). This validation ensures the accuracy of our function's behavior at crucial points.