Problem 25 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 25 through 28, find the absolute maximum and the absolute minimum values (if any) of the given function on the specified interval. $$ f(x)=-2 x^{3}+3 x^{2}+12 x-5 ;-3 \leq x \leq 3 $$

Step-by-Step Solution

Answer

Absolute maximum: 40 at \( x = -3 \); Absolute minimum: -12 at \( x = -1 \).

Step 1 - Find the derivative of the function

To find the critical points of the function, start by taking the derivative of the function. Given that \( f(x) = -2x^3 + 3x^2 + 12x - 5 \), the derivative is found as follows:\[ f'(x) = \frac{d}{dx}(-2x^3 + 3x^2 + 12x - 5) \]Using basic differentiation rules:\[ f'(x) = -6x^2 + 6x + 12 \]

Step 2 - Find critical points

Set the derivative found in Step 1 to zero and solve for \( x \):\[ -6x^2 + 6x + 12 = 0 \]This can be solved using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -6 \), \( b = 6 \), and \( c = 12 \):\[ x = \frac{-6 \pm \sqrt{6^2 - 4(-6)(12)}}{2(-6)} \]\[ x = \frac{-6 \pm \sqrt{36 + 288}}{-12} \]\[ x = \frac{-6 \pm 18}{-12} \]This gives two solutions:\( x = 2 \) and \( x = -1 \)

Step 3 - Evaluate the function at the critical points and endpoints

Next, evaluate the function \( f(x) \) at the critical points found in Step 2 and at the endpoints of the interval, \( x = -3 \) and \( x = 3 \):For \( x = 2 \):\[ f(2) = -2(2)^3 + 3(2)^2 + 12(2) - 5 \]\[ f(2) = -16 + 12 + 24 - 5 = 15 \]For \( x = -1 \):\[ f(-1) = -2(-1)^3 + 3(-1)^2 + 12(-1) - 5 \]\[ f(-1) = 2 + 3 - 12 - 5 = -12 \]For \( x = -3 \):\[ f(-3) = -2(-3)^3 + 3(-3)^2 + 12(-3) - 5 \]\[ f(-3) = 54 + 27 - 36 - 5 = 40 \]For \( x = 3 \):\[ f(3) = -2(3)^3 + 3(3)^2 + 12(3) - 5 \]\[ f(3) = -54 + 27 + 36 - 5 = 4 \]

Step 4 - Determine the absolute maximum and minimum values

Compare the function values obtained in Step 3:\( f(-3) = 40 \)\( f(3) = 4 \)\( f(2) = 15 \)\( f(-1) = -12 \)The absolute maximum value is \( 40 \) at \( x = -3 \) and the absolute minimum value is \( -12 \) at \( x = -1 \).

Key Concepts

Critical Points

Critical points are where a function's derivative equals zero or doesn't exist. These points are crucial because they can indicate where local maxima or minima may occur.
To find critical points, you first need to take the derivative of the function. For the function \( f(x) = -2x^3 + 3x^2 + 12x - 5 \), its derivative is \( f'(x) = -6x^2 + 6x + 12 \).
Next, you set the derivative equal to zero and solve for \( x \):

• \( -6x^2 + 6x + 12 = 0 \)

This quadratic equation can then be solved using various methods, like factoring or the quadratic formula, to find the values of \( x \) where the critical points occur. In this example, solving the equation yields the critical points \( x = 2 \) and \( x = -1 \). These are potential candidates for local extreme values of the function.

Derivative

A derivative represents the rate of change of a function concerning one of its variables. In simpler terms, it can be thought of as the slope of the function at any given point.
To find the derivative of a polynomial function like \( f(x) = -2x^3 + 3x^2 + 12x - 5 \), you apply the power rule: ⥤ For each term \( ax^n \), the derivative is \( anx^{n-1} \).
Using this rule, we get:

• \( \frac{d}{dx}(-2x^3) = -6x^2 \)
• \( \frac{d}{dx}(3x^2) = 6x \)
• \( \frac{d}{dx}(12x) = 12 \)
• The constant term \( -5 \) vanishes as its derivative is zero.

Combining all these, we find the derivative of the function: \( f'(x) = -6x^2 + 6x + 12 \). Knowing how to find and interpret derivatives is essential for identifying critical points and understanding the behavior of the original function.

Quadratic Formula

The quadratic formula is a valuable tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). The formula is expressed as:
\[ x = \frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a} \]
Suppose we have the quadratic equation from the derivative: \( -6x^2 + 6x + 12 = 0 \).
Here, \( a = -6 \), \( b = 6 \), and \( c = 12 \). Plugging these values into the quadratic formula, we get:

• \( x = \frac{-6 \, \pm \, \sqrt{6^2 - 4(-6)(12)}}{2(-6)} \)

Simplifying inside the square root and then outside it gives us:

• \( x = \frac{-6 \, \pm \, 18}{-12} \)
• This results in two solutions: \( x = 2 \) and \( x = -1 \).

These solutions are the critical points where the function changes its slope and can potentially have local maxima or minima.

Interval Evaluation

Evaluating the function within a given interval is key to finding absolute maxima and minima. To do this, you need to evaluate the function at critical points and endpoints of the interval.
For the given function \( f(x) = -2x^3 + 3x^2 + 12x - 5 \) on the interval \( -3 \leq x \leq 3 \), we already found the critical points: \( x = 2 \) and \( x = -1 \). Now we evaluate the function at these points and at the endpoints \( x = -3 \) and \( x = 3 \).

• At \( x = 2 \): \( f(2) = 15 \)
• At \( x = -1 \): \( f(-1) = -12 \)
• At \( x = -3 \): \( f(-3) = 40 \)
• At \( x = 3 \): \( f(3) = 4 \)

Comparing these values, the absolute maximum is \( 40 \) at \( x = -3 \) and the absolute minimum is \( -12 \) at \( x = -1 \). Evaluating the function across the interval ensures that you correctly identify the highest and lowest values within the given range.