Problem 53 - Algebra and Trigonometry

Question

\(51-58\) Graph the polynomial in the given viewing rectangle. Find the coordinates of all local extrema. State each answer rounded to two decimal places. $$ y=x^{3}-12 x+9, \quad[-5,5] \text { by }[-30,30] $$

Step-by-Step Solution

Answer

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

Step 1: Find the First Derivative

To find the local extrema, we first need to find the first derivative of the polynomial function \(y = x^3 - 12x + 9\). The derivative of a function gives us insight into changes in the slope of the tangent line to the graph of the function. The first derivative is calculated as follows: \[y' = \frac{d}{dx}(x^3 - 12x + 9) = 3x^2 - 12.\]

Step 2: Set the First Derivative to Zero

To find the local extrema (i.e., local minima or maxima), set the first derivative equal to zero and solve for \(x\):\[3x^2 - 12 = 0.\]

Step 3: Solve the Equation

Solve the equation \(3x^2 - 12 = 0\) to find the critical points. Divide both sides by 3:\[x^2 - 4 = 0.\]Next, factor the equation:\[(x - 2)(x + 2) = 0.\]Thus, the solutions are:\[x = 2 \text{ and } x = -2.\]

Step 4: Calculate the y-values of the Critical Points

Substitute the critical points \(x = 2\) and \(x = -2\) back into the original function to find their corresponding \(y\)-values:For \(x = 2\):\[y = 2^3 - 12 \cdot 2 + 9 = 8 - 24 + 9 = -7.\]For \(x = -2\):\[y = (-2)^3 - 12 \cdot (-2) + 9 = -8 + 24 + 9 = 25.\]

Step 5: Determine the Type of Extrema Using the Second Derivative

Use the second derivative to determine whether the critical points \(x = 2\) and \(x = -2\) are local minima or maxima. Compute the second derivative:\[y'' = \frac{d^2}{dx^2}(x^3 - 12x + 9) = 6x.\]Evaluate the second derivative at each critical point:At \(x = 2\):\[y''(2) = 6 \cdot 2 = 12.\] Since \(y''(2) > 0\), \(x = 2\) is a local minimum.At \(x = -2\):\[y''(-2) = 6 \cdot (-2) = -12.\] Since \(y''(-2) < 0\), \(x = -2\) is a local maximum.

Step 6: List the Coordinates of the Local Extrema

Based on calculations, the coordinate for the local minimum is \((2, -7)\), and the coordinate for the local maximum is \((-2, 25)\).

Key Concepts

Understanding Local Extrema

Local extrema are the highest or lowest points in a small, surrounding area of a graph. They reveal where a function changes from increasing to decreasing, or vice versa. Local maxima are peaks, while local minima are valleys in the graph. In the graph of a polynomial like \(y = x^3 - 12x + 9\), finding local extrema helps us understand the overall shape and behavior of the curve.

• Local Maximum: A point where the graph changes from rising to falling.
• Local Minimum: A point where the graph shifts from falling to rising.

For the given polynomial, calculating the exact coordinates of local extrema helps in accurately sketching the graph.

Exploring the First Derivative

The first derivative of a function, represented as \(y'\), is key for identifying local extrema. It determines the slope of the tangent line at any point on the graph, providing insights into the graph's steepness and direction. By setting the first derivative to zero, we discover critical points where the potential for a local extremum exists.
For the polynomial \(y = x^3 - 12x + 9\), the first derivative is calculated as \(y' = 3x^2 - 12\). This tells us:

• The function's rate of change at each point along the curve.
• Where the function's rate of change is zero, indicating possible peaks or valleys.

Solving \(3x^2 - 12 = 0\) helps identify \(x = 2\) and \(x = -2\) as critical points.

Insights from the Second Derivative

The second derivative gives us deeper understanding into the graph's curvature and helps ascertain the nature of critical points. It shows whether the graph is concave up (curving upwards) or concave down (curving downwards) at each critical point.
In our polynomial case, the second derivative is \(y'' = 6x\). By evaluating this at our critical points:

• At \(x = 2\), \(y''(2) = 12\) implies the graph is concave up, indicating a local minimum.
• At \(x = -2\), \(y''(-2) = -12\) signifies the graph is concave down, showing a local maximum.

Thus, the second derivative test clarifies whether a critical point is a peak or valley.

Identifying Critical Points

Critical points occur where the first derivative is zero or undefined. These pivotal locations on the graph require further investigation to determine if they represent local maxima, minima, or neither.
For \(y = x^3 - 12x + 9\), critical points were derived by solving \(3x^2 - 12 = 0\), leading us to \(x = 2\) and \(x = -2\). Substituting these values back into the original function provides their corresponding \(y\)-values:

• At \(x = 2\), \(y = -7\), which is a local minimum.
• At \(x = -2\), \(y = 25\), representing a local maximum.

Determining these coordinates gives clarity in sketching the polynomial function's graph, ensuring it reflects the key characteristics accurately.