Problem 20 - Single Variable Calculus

Question

(a) find the intervals on which \(f\) is increasing or decreasing, and (b) find the relative maxima and relative minima of \(\vec{f}\). $$ f(x)=x^{3}(x-6)^{4} $$

Step-by-Step Solution

Answer

The function \(f(x) = x^3(x-6)^4\) is increasing on the intervals \((-\infty, 0)\), \((0, \frac{18}{7})\), and \((6, \infty)\), and decreasing on the interval \((\frac{18}{7}, 6)\). There is a relative maximum at \(x = \frac{18}{7}\), but no relative minimum is found for this function.

Step 1: Finding the first derivative

Calculate the first derivative of \(f(x) = x^3(x-6)^4\) by applying the product rule. The product rule states that \((u \cdot v)' = u' \cdot v + u \cdot v'\). In this case, we have two functions: \(u(x) = x^3\) and \(v(x) = (x-6)^4\). Then, find their derivatives: \(u'(x) = 3x^2\) \(v'(x) = 4(x-6)^3\) Now, apply the product rule: \(f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) = 3x^2(x-6)^4 + x^3 \cdot 4(x-6)^3\) We will use this derivative to find the critical points and the intervals where the function is increasing or decreasing.

Step 2: Finding the critical points

To find the critical points, find the values of \(x\) when \(f'(x) = 0\). In other words, we solve the equation \(f'(x) = 3x^2(x-6)^4 + 4x^3(x-6)^3 = 0\). Factor out the common term \(x^2(x-6)^3\): \(f'(x) = x^2(x-6)^3(3(x-6) + 4x) = x^2(x-6)^3(3x - 18 + 4x) = x^2(x-6)^3(7x - 18)\) This expression is equal to 0 when: 1. \(x = 0\) 2. \(x = 6\) 3. \(x = \frac{18}{7}\) These three values for \(x\) are our critical points. Next, we will find where the function is increasing or decreasing and the relative maxima and minima.

Step 3: Intervals of increasing and decreasing

To determine the intervals of increasing and decreasing, we use the first derivative test. Test points within the intervals formed by the critical points in the first derivative: 1. Choose \(x = -1\), which is in the interval \((-\infty, 0)\). Evaluate, \(f'(-1) = (-1)^2(-7)^3(7(-1) - 18) = (-7)^3(-25) > 0\). Therefore, \(f(x)\) is increasing in this interval. 2. Choose \(x = 3\), in the interval \((0, \frac{18}{7})\). We get \(f'(3) = (3)^2(-3)^3(7(3) - 18) = (-3)^3(3) > 0\). Hence, \(f(x)\) is increasing in this interval as well. 3. Choose \(x = 6.5\), which belongs to the interval \((\frac{18}{7}, 6)\), to find out if \(f'(6.5) > 0\). We notice that \(f'(6.5) = (6.5)^2(-0.5)^3(7(6.5) - 18) = (-0.5)^3(27.5) < 0\). Consequently, \(f(x)\) is decreasing in this range. 4. Choose \(x = 8\), which is within the interval \((6, \infty)\), and calculate \(f'(8) = (8)^2(2)^3(7(8) - 18) = (2)^3(38) > 0\). Therefore, \(f(x)\) is increasing in this interval. We now have the intervals of increase and decrease for \(f(x)\).

Step 4: Finding the relative maxima and minima

By analyzing the intervals of increase and decrease and the critical points, we can identify the relative maxima and minima: - \(x = \frac{18}{7}\): Since \(f(x)\) switches from increasing to decreasing at \(x = \frac{18}{7}\), this point is a relative maximum. Thus, \((\frac{18}{7}, f(\frac{18}{7}))\) is a relative maxima. No relative minima are found in this function. To summarize, the function \(f(x) = x^3(x-6)^4\) has the following properties: 1. Increasing intervals: \((-\infty, 0)\), \((0, \frac{18}{7})\), and \((6, \infty)\). 2. Decreasing interval: \((\frac{18}{7}, 6)\). 3. Relative maxima: \((\frac{18}{7}, f(\frac{18}{7}))\) 4. No relative minima.

Key Concepts

Critical Points

In calculus, critical points are where the first derivative of a function is either zero or undefined. These points are important because they indicate where the function might have a local maximum, minimum, or a point of inflection.

To find them, you need to:

• Calculate the first derivative of the function.
• Set the derivative equal to zero and solve for the variable, often denoted as \(x\).

In our example, given the function \(f(x) = x^3(x-6)^4\), the critical points were found by solving \(f'(x) = 0\). This led to:

• \(x = 0\)
• \(x = 6\)
• \(x = \frac{18}{7}\)

Understanding critical points helps you analyze and understand the behaviour of functions by showing where these significant changes might occur.

Increasing and Decreasing Functions

A key part of analyzing functions is determining where they are increasing or decreasing. This helps us understand the behaviour and shape of the graph of the function.

To determine these intervals:

• Use the first derivative: If \(f'(x) > 0\), the function is increasing.
• If \(f'(x) < 0\), the function is decreasing.

For \(f(x) = x^3(x-6)^4\), we found:

• Increasing: \((-\infty, 0)\), \((0, \frac{18}{7})\), and \((6, \infty)\)
• Decreasing: \((\frac{18}{7}, 6)\)

These intervals tell you how the function behaves over different ranges and help identify potential maxima and minima.

Product Rule

The product rule is a fundamental tool in calculus for finding the derivative of a product of two functions. If you have \(u(x)\) and \(v(x)\), the product rule states:

• \((u \cdot v)' = u' \cdot v + u \cdot v'\)

This means you take the derivative of each function separately and then combine them.

For our function \(f(x) = x^3(x-6)^4\), let:

• \(u(x) = x^3\)
• \(v(x) = (x-6)^4\)

Their derivatives are:

• \(u'(x) = 3x^2\)
• \(v'(x) = 4(x-6)^3\)

Applying the product rule gives us: \[f'(x) = 3x^2(x-6)^4 + x^3 \cdot 4(x-6)^3\] This first derivative is crucial for finding the critical points and understanding where the function is increasing or decreasing.

Relative Extrema

Relative extrema, or local maxima and minima, are the highest or lowest points within a specific interval of a function. These points are found at critical points and are crucial for analyzing the graph's shape.

To determine whether a critical point is a relative maximum or minimum, use the first derivative test:

• If \(f'(x)\) changes from positive to negative, the point is a relative maximum.
• If \(f'(x)\) changes from negative to positive, the point is a relative minimum.

For \(f(x) = x^3(x-6)^4\):

• \(x = \frac{18}{7}\) was identified as a relative maximum because \(f(x)\) changes from increasing to decreasing.

In this case, no relative minimum exists. Identifying relative extrema is essential for understanding the financial, physical, and graphical impacts of the function's behavior.