Problem 50 - 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. $$ y=2 x^{3}-3 x^{2}-12 x-32,[-5,5] \text { by }[-60,30] $$

Step-by-Step Solution

Answer

Local minimum at (2, -52) and local maximum at (-1, -25).

Step 1: Identify the Derivative

To find the local extrema, we need the derivative, as extrema occur where the derivative is zero or undefined. Let's find the derivative of the function \( y = 2x^3 - 3x^2 - 12x - 32 \). The derivative is:\[ y' = \frac{d}{dx}(2x^3 - 3x^2 - 12x - 32) = 6x^2 - 6x - 12 \]

Step 2: Find Critical Points

Set the derivative equal to zero and solve for \( x \) to find the critical points:\[ 6x^2 - 6x - 12 = 0 \]Divide the entire equation by 6 to simplify:\[ x^2 - x - 2 = 0 \]Factor the quadratic equation:\[ (x-2)(x+1) = 0 \]Set each factor equal to 0:\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]

Step 3: Evaluate the Function at Critical Points

Substitute \( x = 2 \) and \( x = -1 \) into the original function to find the \( y \)-coordinates of the extrema.For \( x = 2 \):\[ y = 2(2)^3 - 3(2)^2 - 12(2) - 32 = 16 - 12 - 24 - 32 = -52 \]For \( x = -1 \):\[ y = 2(-1)^3 - 3(-1)^2 - 12(-1) - 32 = -2 - 3 + 12 - 32 = -25 \]

Step 4: Determine the Type of Extrema

To determine whether each critical point is a local maximum or minimum, use the second derivative test.Find the second derivative of \( y \):\[ y'' = 12x - 6 \]Evaluate the second derivative at the critical points:For \( x = 2 \):\[ y''(2) = 12(2) - 6 = 24 - 6 = 18 \]Since \( y''(2) > 0 \), \( x = 2 \) is a local minimum.For \( x = -1 \):\[ y''(-1) = 12(-1) - 6 = -12 - 6 = -18 \]Since \( y''(-1) < 0 \), \( x = -1 \) is a local maximum.

Step 5: State the Local Extrema Coordinates

The local extrema of the polynomial function \( y = 2x^3 - 3x^2 - 12x - 32 \) are as follows:Local minimum at \( (2, -52) \).Local maximum at \( (-1, -25) \). Ensure all answers are stated to two decimal places even though they naturally resulted in integer forms here.

Key Concepts

Local Extrema

Local extrema refer to the points on a graph where a function reaches either a local minimum or a local maximum. In simpler terms, these are like the peaks and valleys we see on a graph within a specific range.
To find local extrema, we rely heavily on derivatives. This is because the slope of the tangent line at these peaks or valleys is always zero, meaning the derivative at these points equals zero.
In the given polynomial function \[ y = 2x^3 - 3x^2 - 12x - 32 \] we calculated the derivative, set it to zero, and found the critical points. These critical points undergo further testing to determine if they are local minima or maxima.

Derivative

A derivative is a fundamental concept in calculus used to measure how a function changes at any given point, much like speedometers measure how fast or slow a car is moving at a particular instant.
Calculating the derivative of our given function \[ y = 2x^3 - 3x^2 - 12x - 32 \] resulted in\[ y' = 6x^2 - 6x - 12 \]. This expression represents the rate of change of the original function, offering insights into its behavior.- **Crucial Insight**: The derivative tells us where the function’s rate of change becomes zero (critical points), which are potential local extrema.- **Next Step**: Set this derivative equal to zero to find points where the slope of the tangent touches zero, indicating a potential turn in the graph’s direction.

Critical Points

Critical points are essentially the heart of finding local extrema. These are the points where the derivative of the function is zero or undefined. In the context of our exercise, we set the derivative\[ y' = 6x^2 - 6x - 12 \] equal to zero to solve for the values of \( x \) that indicate these critical points. - **Solving the Equation**: By simplifying, we get\[ x^2 - x - 2 = 0 \]. Factorizing gives us the solutions\[ (x - 2)(x + 1) = 0 \] which translates to critical points at \( x = 2 \) and \( x = -1 \).- **Importance**: These \( x \) values are crucial since they specify where the function's graph might switch from increasing to decreasing or vice versa, pointing out possible locations of local minima or maxima.

Quadratic Factorization

Quadratic factorization is the process of breaking down a quadratic equation into simpler expressions. For our function's derivative\[ y' = 6x^2 - 6x - 12 \] simplification led us to solve the quadratic equation\[ x^2 - x - 2 = 0 \]. By factorizing it, we identified the solution as:- **Step-by-Step**: - Split the terms: \[ (x - 2)(x + 1) = 0 \] - This results in two straightforward equations \[ x - 2 = 0 \] and \[ x + 1 = 0 \] - Solutions are found quickly as \( x = 2 \) and \( x = -1 \).- **Factorization Benefits**: This process simplifies solving for \( x \), giving us the critical points needed to inspect further for local extrema.