Problem 14 - Thomas Calculus in SI Units

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=1-9 x-6 x^{2}-x^{3}$$

Step-by-Step Solution

Answer

Local extrema: (-1, 5) (max) and (-3, 1) (min); inflection point: (-2, 3).

Step 1: Find the First Derivative

To find the local extreme points of the function, we first need to calculate the first derivative. The derivative of the given function \( y = 1 - 9x - 6x^2 - x^3 \) is \( y' = -9 - 12x - 3x^2 \).

Step 2: Solve for Critical Points

Set the first derivative equal to zero to find the critical points: \( -9 - 12x - 3x^2 = 0 \). This simplifies to \( 3x^2 + 12x + 9 = 0 \). Solve this quadratic equation using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 3 \), \( b = 12 \), and \( c = 9 \). This results in \( x = -1 \) and \( x = -3 \).

Step 3: Determine Local Extrema

Substitute the critical points back into the original function to find the y-coordinates. For \( x = -1 \), \( y(-1) = 1 - 9(-1) - 6(-1)^2 - (-1)^3 = 1 + 9 - 6 + 1 = 5 \). For \( x = -3 \), \( y(-3) = 1 - 9(-3) - 6(-3)^2 - (-3)^3 = 1 + 27 - 54 + 27 = 1 \). Thus, the local extrema are (-1, 5) and (-3, 1). Determine their nature using the second derivative.

Step 4: Find the Second Derivative

Calculate the second derivative to analyze the curvature of the function. The second derivative of \( y \) is \( y'' = -12 - 6x \).

Step 5: Use the Second Derivative Test

Evaluate the second derivative at the critical points. For \( x = -1 \), \( y''(-1) = -12 - 6(-1) = -6 \) (negative suggests a local maximum). For \( x = -3 \), \( y''(-3) = -12 - 6(-3) = 6 \) (positive suggests a local minimum).

Step 6: Find Inflection Points

Set the second derivative equal to zero to find inflection points: \( -12 - 6x = 0 \). Solve for \( x \): \( x = -2 \). Substitute into the original function to find the y-coordinate: \( y(-2) = 1 - 9(-2) - 6(-2)^2 - (-2)^3 = 1 + 18 - 24 + 8 = 3 \). So, the inflection point is (-2, 3).

Step 7: Graph the Function

Plot the function \( y = 1 - 9x - 6x^2 - x^3 \) including local maximum at (-1, 5), local minimum at (-3, 1), and inflection point at (-2, 3). The graph will help visualize the shape and confirm these points.

Key Concepts

Local Extrema

Local extrema refer to the highest or lowest points in a specific section of a curve. These are points where the function changes direction, indicating a possible maximum or minimum value.
To find local extrema, we often utilize derivatives. The first derivative tells us about the slope of the curve:

• If the derivative transitions from positive to negative, we have a local maximum.
• If it transitions from negative to positive, we have a local minimum.

In our example, the critical points (places where the derivative equals zero or doesn't exist) were at \( x = -1 \) and \( x = -3 \). By evaluating these in the second derivative, we discovered a local maximum at \( (-1, 5) \) and a local minimum at \( (-3, 1) \). This indicates where the curve peaks and dips respectively. The second derivative helps confirm the nature:

• A positive second derivative indicates concave up (local minimum).
• A negative second derivative suggests concave down (local maximum).

Remember, visualizing these on a graph can be very helpful to clearly see the valleys and hills of the function.

Inflection Points

Inflection points are intriguing because they are where the curve changes its concavity. In simpler terms, this is where a curve goes from curving upward (like a frown) to curving downward (like a smile), or vice versa.
Inflection points happen when the second derivative of a function is zero or undefined. This change indicates that the curve's rate of change of slope itself alters:

• An inflection point doesn’t necessarily mean a maximum or minimum, but a turning point in the shape.

For our function, setting the second derivative \( y'' = -12 - 6x \) to zero gives us \( x = -2 \). Replacing \( x = -2 \) back into the original function, we find that the inflection point is at \( (-2, 3) \). This subtle inflection indicates how the curve morphs and can greatly aid in sketching a precise graph. Recognizing inflection points helps us understand the underlying dynamics of the function and make better predictions about future behavior.

Critical Points

Critical points are crucial parts of calculus, acting as the primary checkpoints for analyzing a function's behavior. They're found by setting the first derivative of a function to zero, as they indicate spots where the tangent to the curve is horizontal.
These areas are essential when looking for potential local extrema. Critical points mark where major changes in the function's direction might occur.

• If the function is smooth and continuous, critical points can expose the highest or lowest points of a curve.

In our task with the function \( y = 1 - 9x - 6x^2 - x^3 \), solving \( y' = -9 - 12x - 3x^2 = 0 \) provided the critical points \( x = -1 \) and \( x = -3 \). These critical points serve as the investigative starting points of understanding the more detailed behavior of the function. Once identified, these points can be explored further using the second derivative to reveal more about the function’s nature—helping confirm whether they're maxima, minima, or neither.