Problem 224 - Calculus Volume 1

Question

For the following exercises, determine $$ \begin{array}{l}{\text { a. intervals where } f \text { is increasing or decreasing, }} \\ {\text { b. local minima and maxima of } f \text { , }} \\\ {\text { c. intervals where } f \text { is concave up and concave }} \\\ {\text { down, and }} \\ {\text { d. the inflection points of } f \text { . }}\end{array} $$ $$ f(x)=x^{2}-6 x $$

Step-by-Step Solution

Answer

The function is decreasing on \((-\infty, 3)\), increasing on \((3, \infty)\), and concave up on \((-\infty, \infty)\). Local minimum at \((3, -9)\). No inflection points.

Step 1: Find the First Derivative

To determine the intervals where the function is increasing or decreasing, we need to calculate the first derivative of the function. Given, \( f(x) = x^2 - 6x \), the first derivative is \( f'(x) = 2x - 6 \).

Step 2: Determine Critical Points

Set the first derivative equal to zero and solve for \( x \): \[ 2x - 6 = 0 \] \[ 2x = 6 \] \[ x = 3 \] Critical points are where the derivative is zero or undefined.

Step 3: Test Intervals Around Critical Points

Determine the sign of \( f'(x) \) in the intervals created by the critical points. Our critical point is \( x = 3 \). We choose test points from the intervals \((-\infty, 3)\) and \((3, \infty)\).- For \( x = 2 \), \( f'(2) = 2(2) - 6 = -2 \) (decreasing).- For \( x = 4 \), \( f'(4) = 2(4) - 6 = 2 \) (increasing).

Step 4: Find Local Minima and Maxima

Since the function changes from decreasing to increasing at \( x = 3 \), we have a local minimum at this point. The y-value at \( x = 3 \) is \( f(3) = 3^2 - 6(3) = -9 \). Thus, the local minimum is \((3, -9)\).

Step 5: Find the Second Derivative

To determine concavity, find the second derivative of \( f(x) \). \( f''(x) = 2 \).

Step 6: Determine Intervals of Concavity

Since the second derivative \( f''(x) = 2 \) is positive for all \( x \), this implies the function is concave up on the entire domain \((-\infty, \infty)\). However, there are no intervals where \( f(x) \) is concave down.

Step 7: Identify Inflection Points

Inflection points occur where the concavity changes, which is determined by the points where \( f''(x) = 0 \). Since \( f''(x) = 2 \) and never zero, there are no inflection points.

Key Concepts

First Derivative Test

The First Derivative Test is an essential tool in calculus for determining where a function is increasing or decreasing. To use this test, we first find the derivative of a function. For example, if we have the function \( f(x) = x^2 - 6x \), the first derivative is calculated to be \( f'(x) = 2x - 6 \).
Once we have the derivative, we find the critical points by setting \( f'(x) = 0 \) and solving for \( x \). In this case, doing so yields \( x = 3 \).
Next, we test intervals around the critical points to determine the sign of \( f'(x) \):

• If \( f'(x) > 0 \), the function is increasing on that interval.
• If \( f'(x) < 0 \), the function is decreasing on that interval.

Using test points, we find \( f'(x) \) is negative on \((-\infty, 3)\) and positive on \((3, \infty)\), indicating the function decreases before \( x = 3 \) and increases after. Therefore, using the First Derivative Test, we conclude that the function has a local minimum at \( x = 3 \).

Critical Points

Critical points are where the derivative of a function is either zero or undefined. These points are crucial for analyzing the behavior of a function, such as determining local maxima and minima.
In our example, \( f(x) = x^2 - 6x \), the first derivative \( f'(x) = 2x - 6 \) reveals one critical point at \( x = 3 \) because this is where the derivative equals zero.
Critical points can help identify important features of a graph:

• Potential locations for local maxima or minima.
• Change of a function's increasing or decreasing behavior.

Once a critical point is identified, further tests like the First or Second Derivative Test can determine the nature of these points.

Concavity

Concavity describes how the graph of a function curves along its path. To analyze concavity, the second derivative of the function is essential. For our function \( f(x) = x^2 - 6x \), the second derivative is \( f''(x) = 2 \).
Concavity is understood through the signs of the second derivative:

• If \( f''(x) > 0 \), the graph is concave up, resembling a U-shape.
• If \( f''(x) < 0 \), the graph is concave down, resembling an upside-down U.

Since \( f''(x) = 2 \) is positive for all \( x \), the function is concave up over the entire domain \((-\infty, \infty)\). This uniform concavity indicates a consistent curve upwards, and there are no intervals where \( f(x) \) is concave down.

Inflection Point

An inflection point occurs where a function's concavity changes from up to down or down to up. Typically, this occurs where the second derivative equals zero or changes sign.
For the function \( f(x) = x^2 - 6x \), we calculate the second derivative, which is \( f''(x) = 2 \). Notably, since \( f''(x) \) is always positive and constant, it never equals zero, and there are no sign changes.
Thus, in this case, there are no inflection points. Remember that the lack of inflection points is associated with functions that are consistently curved in a single direction, whether concave up or down, across their entire domain.