Problem 12 - Thomas Calculus in SI Units

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=x(6-2 x)^{2}$$

Step-by-Step Solution

Answer

Local max at \((1, 16)\), local min at \((3, 0)\), inflection point at \((2, 8)\).

Step 1: Differentiate the Function

First, we need to find the derivative of the function to identify critical points. The function is given as \( y = x (6 - 2x)^2 \). We use the product rule for differentiation, \( (uv)' = u'v + uv' \), where \( u = x \) and \( v = (6 - 2x)^2 \). Thus, we find: 1. \( u' = 1 \) and \( v = (6 - 2x)^2 \). 2. \( v' = 2(6 - 2x)(-2) = -4(6 - 2x) \). Applying the product rule: \[ y' = 1 \cdot (6 - 2x)^2 + x \cdot -4(6 - 2x) \] \[ = (6 - 2x)^2 - 4x(6 - 2x) \] \[ = (6 - 2x)^2 - 24x + 8x^2 \].

Step 2: Simplify and Set the Derivative to Zero

To find the critical points, we set the derivative \( y' \) to zero. Thus, simplify the expression: \[ y' = 36 - 24x + 4x^2 - 24x + 8x^2 \] \[ = 4x^2 + 8x^2 - 48x + 36 \] \[ = 12x^2 - 48x + 36 \]. Solving \( y' = 0 \): Divide through by 12: \[ x^2 - 4x + 3 = 0 \] Factor the quadratic equation: \[ (x - 3)(x - 1) = 0 \]. Thus, \( x = 3 \) and \( x = 1 \) are critical points.

Step 3: Use the Second Derivative Test

Calculate the second derivative to determine the nature of the critical points. Start from: \( y' = 12x^2 - 48x + 36 \). Differentiate again: \[ y'' = (12x^2 - 48x + 36)' = 24x - 48 \]. Evaluate \( y'' \) at the critical points: For \( x = 1 \): \[ y''(1) = 24(1) - 48 = -24 \] (local maximum). For \( x = 3 \): \[ y''(3) = 24(3) - 48 = 24 \] (local minimum).

Step 4: Identify Inflection Points

To find inflection points, set the second derivative \( y'' \) to zero. Solve \( 24x - 48 = 0 \): \[ 24x = 48 \] \[ x = 2 \]. Check a test point to see if concavity changes around \( x = 2 \): For \( x < 2 \), \( y'' < 0 \); for \( x > 2 \), \( y'' > 0 \). Thus, an inflection point occurs at \( x = 2 \). The coordinates can be determined by substituting \( x = 2 \) back into the original equation \( y = x(6 - 2x)^2 \): \[ y = 2(6 - 4)^2 = 2 \cdot 4 = 8 \]. Therefore, the inflection point is \((2, 8)\).

Step 5: Identify Absolute Extreme Points

Evaluate the function at the critical points and endpoints (if specified) to determine absolute extrema. - \( y(1) = 1(6 - 2(1))^2 = 16 \), which is a local maximum. - \( y(3) = 3(6 - 6)^2 = 0 \), which is a local minimum. If considering within a specific domain, compare these to the values at endpoints. Here, \( 1 \) is the highest value identified as an absolute maximum, and \( 3 \) is the lowest as an absolute minimum.

Step 6: Graphing the Function

Plot the function \( y = x(6 - 2x)^2 \) to verify graphical interpretation:1. Plot points at critical points: \((1, 16)\) and \((3, 0)\).2. Plot the inflection point: \((2, 8)\).3. The graph confirms that these are extrema and inflection points: - Observe the descent from a peak at \( x = 1 \), - A curvature switch at \( x = 2 \), and - An increasing trend after \( x = 3 \). Sketch the behavior between these points.

Key Concepts

Product Rule

The product rule is an essential tool in calculus for differentiating expressions where two functions are multiplied together. This rule is represented as \((uv)' = u'v + uv'\). Here, \(u\) and \(v\) are functions of \(x\). In the original exercise function \( y = x(6-2x)^2 \), \( u = x \) and \( v = (6-2x)^2 \).

• First, differentiate \( u \), giving \( u' = 1 \), since the derivative of \( x \) is \( 1 \).
• Next, differentiate \( v \). Consider \( v = (6 - 2x)^2 \), which requires the chain rule for differentiation, resulting in \( v' = -4(6 - 2x) \).

After differentiating, apply the product rule:

• \( y' = 1 \cdot (6 - 2x)^2 + x \cdot (-4(6 - 2x)) \)

Simplify this expression to help find the critical points.
The beauty of the product rule lies in its ability to handle complex functions that are multiplied, breaking them down into more manageable parts for differentiation.

Critical Points

Critical points in calculus occur where the derivative of a function is zero or undefined. They are key indicators of potential maxima or minima on the graph of a function. To find these points for the function \( y = x(6 - 2x)^2 \), set the derivative equal to zero.
Starting from the derivative found with the product rule:

• \( y' = (6 - 2x)^2 - 24x + 8x^2 \)

Simplify and solve:

• \( 12x^2 - 48x + 36 = 0 \)

Divide the equation by 12 to make simplifying easier:

• \( x^2 - 4x + 3 = 0 \)

By factoring the quadratic equation, we find:

• \((x - 3)(x - 1) = 0 \)

Hence, the solutions are \( x = 3 \) and \( x = 1 \).
These critical points are where changes in the direction of the graph might occur, and further testing will determine their nature.

Second Derivative Test

The second derivative test provides a method to classify each critical point found from the first derivative as either a local maximum or minimum. It involves taking the second derivative \( y'' \) of the original function and evaluating it at each critical point.
For the function \( y = x(6 - 2x)^2 \), the second derivative is:

• \( y'' = 24x - 48 \)

Evaluate \( y'' \) at the critical points:

• At \( x = 1 \), \( y''(1) = 24(1) - 48 = -24 \), indicating a local maximum.
• At \( x = 3 \), \( y''(3) = 24(3) - 48 = 24 \), indicating a local minimum.

The sign of \( y'' \) at a critical point tells us:

• If \( y'' > 0 \), the function has a local minimum there (concave up).
• If \( y'' < 0 \), the function has a local maximum (concave down).

This test uses the geometric interpretation of the second derivative helping us understand the curve's shape at critical points.

Inflection Points

Inflection points occur where the curvature of a graph changes sign, from concave up to concave down, or vice versa. They are identified using the second derivative \( y'' \). To locate inflection points for the function \( y = x(6 - 2x)^2 \), set the second derivative equal to zero.
Using the second derivative found:

• \( 24x - 48 = 0 \)

Solving gives:

• \( x = 2 \)

Plugging \( x = 2 \) back into the original function gives us \( y = 8 \), resulting in the inflection point \((2, 8)\).
To confirm this, test intervals around \( x = 2 \):

• For \( x < 2 \), if \( y'' < 0 \), and for \( x > 2 \), if \( y'' > 0 \), then an inflection point exists.

Inflection points indicate changes in the concavity of the curve, providing key insight into the graph's behavior beyond what maxima and minima can tell us.