Problem 4 - Calculus: Early Transcendentals

Question

Find all critical points and identify them as local maximum points, local minimum points, or neither. $$ y=x^{3}-9 x^{2}+24 x $$

Step-by-Step Solution

Answer

The critical points are at \( x = 2 \) (local maximum) and \( x = 4 \) (local minimum).

Step 1: Find the derivative

To find the critical points of the function, we start by finding its derivative, which will allow us to locate where the slope is zero or undefined. The function is given by \( y = x^3 - 9x^2 + 24x \). Differentiating with respect to \( x \), we get: \[ y' = 3x^2 - 18x + 24. \]

Step 2: Set the derivative to zero

Critical points occur where the first derivative is zero or undefined. Since the derivative is a polynomial, it is defined for all \( x \). So, we need to solve for \( x \) in \[ 3x^2 - 18x + 24 = 0. \]

Step 3: Solve the quadratic equation

Solve the quadratic equation \( 3x^2 - 18x + 24 = 0 \) using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -18 \), \( c = 24 \). Pre-calculate the discriminant: \( (-18)^2 - 4(3)(24) = 324 - 288 = 36 \). Thus, \[ x = \frac{18 \pm \sqrt{36}}{6} = \frac{18 \pm 6}{6}. \] This simplifies to \( x = 4 \) and \( x = 2 \).

Step 4: Determine the nature of critical points

Now use the second derivative test to determine the nature of the critical points. First, find the second derivative: \[ y'' = 6x - 18 \]. Evaluate \( y'' \) at each critical point. Substitute \( x = 2 \) into \( y'' \): \( y''(2) = 6(2) - 18 = -6 \), which is less than zero, indicating a local maximum. Substitute \( x = 4 \) into \( y'' \): \( y''(4) = 6(4) - 18 = 6 \), which is greater than zero, indicating a local minimum.

Key Concepts

Derivative

Let's start with understanding the concept of a derivative. The derivative of a function represents the rate at which the function is changing at any given point. Think of it as the slope of the tangent line to the graph at a specific point. This can tell us how the function is increasing or decreasing at that point.

For the function given in the exercise, the derivative is calculated as follows: if we have a function \( y = x^3 - 9x^2 + 24x \), we find its derivative by applying the power rule:

• Differentiate each term separately: \( \frac{d}{dx}(x^3) = 3x^2 \)
• \(\frac{d}{dx}(-9x^2) = -18x \)
• \(\frac{d}{dx}(24x) = 24 \)

Thus, the derivative is given by \( y' = 3x^2 - 18x + 24 \).

Finding where this derivative equals zero helps us locate critical points, where the function could have a local maximum or minimum.

Quadratic Equation

The second core concept is solving a quadratic equation to find the critical points. A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \).
The solution method used here is the quadratic formula, expressed as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this scenario, the critical points stem from the quadratic equation that arises from setting the derivative to zero:\[ 3x^2 - 18x + 24 = 0 \].
Applying the quadratic formula, we first find the discriminant:

• Calculate \( (-18)^2 - 4(3)(24) = 324 - 288 = 36 \)
• Then use \( x = \frac{18 \pm 6}{6} \)
• This simplifies to bounding values \( x = 4 \) and \( x = 2 \)

These solutions represent potential locations where the critical points of the function are found.

Second Derivative Test

The second derivative test is used to classify the critical points obtained from the first derivative. Essentially, it helps determine whether each critical point is a local maximum, minimum, or neither. The test checks the concavity of the function at the critical points.

To perform the second derivative test, calculate the second derivative of the function. The second derivative, denoted by \( y'' \), is the derivative of the derivative:\[ y'' = 6x - 18 \].

Evaluate this second derivative at each critical point:

• If \( y''(x) > 0 \), the graph is concave up at that point, indicating a local minimum.
• If \( y''(x) < 0 \), the graph is concave down, indicating a local maximum.
• If \( y''(x) = 0 \), the test is inconclusive.

This helps in understanding the nature of each critical point found from the first derivative.

Local Maximum

A local maximum is a point on the graph where the function reaches a peak higher than its surrounding points. At this point, the function changes direction from increasing to decreasing. In our exercise, use the second derivative to find where \( y'' \) is less than zero.

Checking the critical point when \( x = 2 \):

• Substitute into the second derivative: \( y''(2) = 6(2) - 18 = -6 \)
• Since \( y''(2) < 0 \), it indicates the graph is concave down, confirming that \( x = 2 \) is a local maximum.

This clear understanding helps identify where your function results in a peak locally on the graph.

Local Minimum

A local minimum is a point where the function reaches a valley lower than any nearby points. At a local minimum, the function changes from decreasing to increasing. From our evaluation, we determine this using the second derivative test.

For the critical point found at \( x = 4 \):

• Check: \( y''(4) = 6(4) - 18 = 6 \)
• Since \( y''(4) > 0 \), the function is concave up, which indicates a local minimum.

Identifying such points helps in understanding the dips of the function graphically, aiding in complete analysis of its behavior around that region.