Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

Identify the critical points and find the maximum value and minimum value on the given interval. $$ G(x)=\frac{1}{5}\left(2 x^{3}+3 x^{2}-12 x\right) ; I=[-3,3] $$

Step-by-Step Solution

Answer

The maximum value is 9 at \(x = 3\) and the minimum value is -1.4 at \(x = 1\).

Step 1: Find the Derivative

To find the critical points, we need the first derivative of the function. Let's differentiate the given function: \[ G(x) = \frac{1}{5}(2x^3 + 3x^2 - 12x) \] Derivate each term: \[ G'(x) = \frac{1}{5}(6x^2 + 6x - 12) = \frac{6}{5}x^2 + \frac{6}{5}x - \frac{12}{5} \]

Step 2: Solve for Critical Points

To find the critical points, we set the derivative equal to zero and solve for \(x\): \[ \frac{6}{5}x^2 + \frac{6}{5}x - \frac{12}{5} = 0 \] Solve this quadratic equation by multiplying through by 5 to clear the fractions: \[ 6x^2 + 6x - 12 = 0 \] Divide by 6: \[ x^2 + x - 2 = 0 \] Factor the equation: \[ (x + 2)(x - 1) = 0 \] Thus, \(x = -2\) and \(x = 1\) are critical points.

Step 3: Evaluate at Critical Points and Endpoints

Evaluate \(G(x)\) at the critical points \(x = -2\) and \(x = 1\), and also at the endpoints of the interval \(I = [-3, 3]\):1. \(G(-3) = \frac{1}{5}(2(-3)^3 + 3(-3)^2 - 12(-3)) = \frac{1}{5}(-54 + 27 + 36) = 1.8\)2. \(G(-2) = \frac{1}{5}(2(-2)^3 + 3(-2)^2 - 12(-2)) = \frac{1}{5}(-16 + 12 + 24) = 4\)3. \(G(1) = \frac{1}{5}(2(1)^3 + 3(1)^2 - 12(1)) = \frac{1}{5}(2 + 3 - 12) = -1.4\)4. \(G(3) = \frac{1}{5}(2(3)^3 + 3(3)^2 - 12(3)) = \frac{1}{5}(54 + 27 - 36) = 9\)

Step 4: Identify Maximum and Minimum Values

By comparing the values found at the critical points and endpoints, we identify the maximum and minimum values:- Maximum value is \(G(3) = 9\),- Minimum value is \(G(1) = -1.4\).

Key Concepts

Derivative Calculation

Derivatives are essential in calculus for finding rates of change. To determine critical points, we start by differentiating the function. Given a function like \[ G(x) = \frac{1}{5}(2x^3 + 3x^2 - 12x), \]we apply the power rule to each term. The power rule states that \( d/dx \{x^n\} = nx^{n-1} \).

• Differentiating \( 2x^3 \) gives \( 6x^2 \).
• Differentiating \( 3x^2 \) gives \( 6x \).
• Differentiating \( -12x \) gives \( -12 \).

Hence, the derivative of the function is:\[ G'(x) = \frac{6}{5}x^2 + \frac{6}{5}x - \frac{12}{5}. \]This derivative helps us locate critical points by setting it equal to zero.

Quadratic Equations

After finding the derivative, solving quadratic equations becomes essential. Quadratics generally follow the form \( ax^2 + bx + c = 0 \). In our function's derivative, \[ \frac{6}{5}x^2 + \frac{6}{5}x - \frac{12}{5} = 0, \]all coefficients are fractions.
This can be simplified by multiplying each term by 5, yielding \[ 6x^2 + 6x - 12 = 0. \]We further simplify by dividing each coefficient by 6:\[ x^2 + x - 2 = 0. \]This is a simple quadratic equation that we can solve by factoring:

• Find two numbers that multiply to -2 and add to 1. These numbers are 2 and -1.
• Thus, the quadratic factors to \( (x + 2)(x - 1) = 0 \).
• Solving each factor for zero gives critical points at \( x = -2 \) and \( x = 1 \).

Function Evaluation

Now that we have the critical points, evaluating the function at specific points gives insight into the behavior of the function. The function is \[ G(x) = \frac{1}{5}(2x^3 + 3x^2 - 12x). \]We evaluate \( G(x) \) at critical points \( x = -2 \) and \( x = 1 \), as well as at the interval endpoints \( x = -3 \) and \( x = 3 \):

• Calculate \( G(-3) \): Substitute \(-3\) into \( G(x) \), resulting in \( G(-3) = 1.8 \).
• Calculate \( G(-2) \): Substitute \(-2\) into \( G(x) \), resulting in \( G(-2) = 4 \).
• Calculate \( G(1) \): Substitute \(1\) into \( G(x) \), resulting in \( G(1) = -1.4 \).
• Calculate \( G(3) \): Substitute \(3\) into \( G(x) \), resulting in \( G(3) = 9 \).

Each value represents the output of the function at those specific points.

Maximum and Minimum Values

The process of identifying maximum and minimum values involves comparing function outputs at critical points and endpoints. A maximum is the highest value in a specified interval, while a minimum is the lowest.
For example, after evaluating the function at \( x = -3, -2, 1, \) and \( 3 \), we have these values:

• \( G(-3) = 1.8 \)
• \( G(-2) = 4 \)
• \( G(1) = -1.4 \)
• \( G(3) = 9 \)

Upon comparison, the maximum value within the interval \([-3, 3]\) is 9 (at \( x = 3 \)), and the minimum value is -1.4 (at \( x = 1 \)).Determining these values helps in understanding how a function behaves, giving insights such as profitability, efficiency, or even physical phenomena in real-world problems.