Problem 39 - Essential Calculus Early Transcendentals

Question

\(37-50=\) Find the absolute maximum and absolute minimum values of \(f\) on the given interval. $$f(x)=2 x^{3}-3 x^{2}-12 x+1, \quad[-2,3]$$

Step-by-Step Solution

Answer

Absolute maximum is 8 at \(x = -1\) and absolute minimum is -19 at \(x = 2\).

Step 1: Find the first derivative

To locate the critical points where the function might have a maximum or minimum, first find the derivative of the function. The given function is \(f(x) = 2x^3 - 3x^2 - 12x + 1\). Differentiating with respect to \(x\), we get:\[ f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 12x + 1) = 6x^2 - 6x - 12 \]

Step 2: Set the derivative to zero and solve

To find the critical points, set the first derivative equal to zero and solve for \(x\):\[ 6x^2 - 6x - 12 = 0 \]Simplify by dividing the entire equation by 6:\[ x^2 - x - 2 = 0 \]Now factor the quadratic equation:\( (x - 2)(x + 1) = 0 \)So, the solutions are \(x = 2\) and \(x = -1\).

Step 3: Evaluate the function at critical points and endpoints

Evaluate \(f(x)\) at the critical points and at the endpoints of the interval \([-2, 3]\):1. \(f(-2) = 2(-2)^3 - 3(-2)^2 - 12(-2) + 1 = -16 - 12 + 24 + 1 = -3\)2. \(f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 1 = -2 - 3 + 12 + 1 = 8\)3. \(f(2) = 2(2)^3 - 3(2)^2 - 12(2) + 1 = 16 - 12 - 24 + 1 = -19\)4. \(f(3) = 2(3)^3 - 3(3)^2 - 12(3) + 1 = 54 - 27 - 36 + 1 = -8\)

Step 4: Identify absolute maximum and minimum

Compare the function values obtained to determine the absolute maximum and minimum:- \(f(-2) = -3\)- \(f(-1) = 8\) (absolute maximum)- \(f(2) = -19\) (absolute minimum)- \(f(3) = -8\)Thus, the absolute maximum value is 8 and occurs at \(x = -1\), and the absolute minimum value is -19 and occurs at \(x = 2\).

Key Concepts

Critical Points

When analyzing a function to find absolute extrema, critical points are key. They are values of \( x \) within an open interval where the first derivative \( f'(x) \) is either zero or undefined. These points are essential as they help us identify where a function may have local maxima or minima, which could be absolute in a given interval.
To locate critical points effectively:

• Calculate the first derivative of the function.
• Set the derivative equal to zero and solve for \( x \).
• Check if the derivative does not exist for certain \( x \) values.

In our specific problem, finding the critical points led us to discover that \( x = 2 \) and \( x = -1 \) were crucial values for further evaluation on the interval \([-2, 3]\). These calculations laid the groundwork for identifying potential extreme values of the function.

First Derivative

The first derivative represents the rate of change or slope of the original function \( f(x) \). It's a fundamental tool in calculus for analyzing a function's behavior. For the polynomial function \( f(x) = 2x^3 - 3x^2 - 12x + 1 \), finding the first derivative gives:\[ f'(x) = 6x^2 - 6x - 12 \]
This derivative helps reveal where the function's slope changes from positive to negative or vice versa, indicating potential peaks or troughs, known as critical points.
To fully understand the function's contours:

• Calculate \( f'(x) \), which gives insight into the function's increasing or decreasing nature.
• Solve \( f'(x) = 0 \) to get critical points.
• Simplify the equation to make solving easier, as done here by dividing through by 6.

These steps allow us to pinpoint critical points which we further scrutinize for absolute maximum or minimum values in the specified interval.

Evaluating Endpoints

When determining absolute extrema, don’t overlook the function's behavior at the endpoints of the interval. By evaluating \( f(x) \) at the interval boundaries, you ensure no extremum is missed.
For the interval \([-2, 3]\), substitute these values into the function:

• Calculate \( f(-2) \)
• Calculate \( f(3) \)

These steps confirm whether the highest or lowest points of \( f(x) \) occur at either \( x = -2 \) or \( x = 3 \).
Remember, absolute extrema can sometimes appear at these edge points, making this evaluation crucial. We found that while the absolute minimum was not at an endpoint, evaluating these positions still confirmed that important feature.

Function Analysis

Function analysis involves a comprehensive evaluation of both the derivative data and the interval boundaries. It combines the findings from critical points and endpoint evaluations to definitively determine absolute extrema.
In our scenario:- Calculated both critical points and endpoints were crucial.- Evaluated \( f(x) \) at \(-2, -1, 2, 3\) to get all possible extremum values.- Comparisons between these values led to identifying the absolute maximum and minimum.
By pulling together all these elements, you ensure that the highest and lowest values across the entire interval are thoroughly and accurately identified, which is key in many real-world applications involving optimization and analysis.