Problem 49 - Algebra and Trigonometry

Question

Graph the polynomial in the given viewing rectangle. Find the coordinates of all local extrema. State each answer correct to two decimal places. $$ y=x^{3}-12 x+9, \quad[-5,5] \text { by }[-30,30] $$

Step-by-Step Solution

Answer

Local minimum at (2, -7) and local maximum at (-2, 25).

Step 1: Calculate the Derivative

To find the local extrema of the polynomial, we need to calculate its derivative and set it equal to zero. The given polynomial is \( y = x^3 - 12x + 9 \). Its derivative is obtained by differentiating:\[\frac{dy}{dx} = 3x^2 - 12\]

Step 2: Solve for Critical Points

Set the derivative equal to zero to find the critical points:\[3x^2 - 12 = 0\]Solve for \( x \):\[3x^2 = 12 \x^2 = 4 \x = \pm 2\]The critical points are \( x = 2 \) and \( x = -2 \).

Step 3: Determine Local Extrema Using Second Derivative Test

Apply the second derivative test to determine the nature of the critical points. First, find the second derivative:\[\frac{d^2y}{dx^2} = 6x\]Evaluate the second derivative at each critical point:1. For \( x = 2 \): \[ \frac{d^2y}{dx^2} = 6(2) = 12 > 0 \] This indicates a local minimum.2. For \( x = -2 \): \[ \frac{d^2y}{dx^2} = 6(-2) = -12 < 0 \] This indicates a local maximum.

Step 4: Calculate the Coordinates of the Local Extrema

Use the original function \( y = x^3 - 12x + 9 \) to find the \( y \)-coordinates of the local extrema:1. When \( x = 2 \): \[ y = (2)^3 - 12(2) + 9 = 8 - 24 + 9 = -7 \] The local minimum is at (2, -7).2. When \( x = -2 \): \[ y = (-2)^3 - 12(-2) + 9 = -8 + 24 + 9 = 25 \] The local maximum is at (-2, 25).

Step 5: Check the Graph in the Given Viewing Rectangle

Draw the graph of \( y = x^3 - 12x + 9 \) within the viewing rectangle \([-5, 5] \) by \([-30, 30]\). The graph will show the polynomial curve passing through the local maximum at (-2, 25), and the local minimum at (2, -7).

Key Concepts

Local Extrema

Local extrema refer to the highest or lowest points on a graph within a specific interval. These points are called local maxima and minima, respectively. They provide valuable insights about the behavior of a polynomial function. In our polynomial, the local maxima and minima tell us where the function reaches its peak and its lowest dip inside the interval we're evaluating.
To find these extrema, follow these steps:

• Identify the function (\( y = x^3 - 12x + 9 \)) you are working with.
• Compute the derivative to find points on the graph where the slope is zero or undefined. These are your potential extrema.
• Perform further tests, like the second derivative test, to confirm whether each point is a maximum or a minimum.

The goal is to use these extrema to better understand the shape and nature of the graph in the given interval, which in this case is \([-5,5]\). They can often help predict the function's behavior beyond where it's explicitly evaluated.

Critical Points

Critical points are essential for understanding the turning points and behavior changes in a polynomial function. These points occur where the derivative of the function equals zero or is undefined. In our example, you start by calculating the first derivative, \( \frac{dy}{dx} = 3x^2 - 12 \).
By setting this derivative equal to zero:

• \( 3x^2 - 12 = 0 \)
• Solve to find: \( x^2 = 4 \)
• This gives us critical points at \( x = 2 \) and \( x = -2 \).

These critical points are crucial because they tell us where to look for local maxima or minima. By exploring these points further using tests such as the second derivative test, we can confirm their nature as turning points in the graph.

Derivative

The derivative of a function expresses how the function value changes as its input changes. For a polynomial function like \( y = x^3 - 12x + 9 \), the derivative is crucial for finding critical points where the slope of the tangent to the curve is zero. This derivative is \( \frac{dy}{dx} = 3x^2 - 12 \).
This process involves:

• Understanding rates of change.
• Applying rules of differentiation to simplify obtaining \( \frac{dy}{dx} \) from the original equation.
• Finding values of \( x \) where the derivative equals zero for critical points.

Remember, knowing the derivative helps not only in finding where peaks and troughs occur but also aids in sketching the curve dynamically.

Second Derivative Test

The second derivative test helps determine if a critical point is a local maximum, minimum, or a saddle point. Once the first derivative is calculated and critical points identified, the second derivative \( \frac{d^2y}{dx^2} \) is derived.
For this polynomial:

• The second derivative is \( \frac{d^2y}{dx^2} = 6x \).
• You evaluate it at each critical point:
  • At \( x = 2 \), \( 6(2) = 12 > 0 \), indicating a local minimum.
  • At \( x = -2 \), \( 6(-2) = -12 < 0 \), indicating a local maximum.

The signs of the second derivative at the critical points tell us about the concavity there - if it's positive, the function is concave up (minimum), and if negative, concave down (maximum). This method provides a clearer picture of the graph's features.