Problem 10 - Calculus Early Transcendentals

Question

Find the absolute maximum and minimum values of \(f\) on the given closed interval, and state where those values occur. \(f(x)=2 x^{3}+3 x^{2}-12 x ;[-3,2]\)

Step-by-Step Solution

Answer

Absolute maximum is 20 at \( x = -2 \), and absolute minimum is -7 at \( x = 1 \).

Step 1: Find the derivative of the function

First, we need to find the first derivative of the function to identify the critical points. The given function is \( f(x) = 2x^3 + 3x^2 - 12x \). We use the power rule to differentiate the function:\( f'(x) = \frac{d}{dx}(2x^3 + 3x^2 - 12x) = 6x^2 + 6x - 12 \).

Step 2: Find the critical points

Set the derivative equal to zero and solve for \(x\) to find the critical points:\( 6x^2 + 6x - 12 = 0 \).Factor the quadratic equation:\( 6(x^2 + x - 2) = 0 \).This simplifies to \( 6(x - 1)(x + 2) = 0 \), giving us critical points at \( x = 1 \) and \( x = -2 \).

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

Evaluate \( f(x) \) at the critical points \( x = 1 \) and \( x = -2 \), and at the endpoints of the interval \( x = -3 \) and \( x = 2 \):- \( f(-3) = 2(-3)^3 + 3(-3)^2 - 12(-3) = -54 + 27 + 36 = 9 \).- \( f(-2) = 2(-2)^3 + 3(-2)^2 - 12(-2) = -16 + 12 + 24 = 20 \).- \( f(1) = 2(1)^3 + 3(1)^2 - 12(1) = 2 + 3 - 12 = -7 \).- \( f(2) = 2(2)^3 + 3(2)^2 - 12(2) = 16 + 12 - 24 = 4 \).

Step 4: Determine the absolute maximum and minimum values

Compare the values obtained in the previous step to determine the absolute maximum and minimum values:- The smallest value is \( -7 \) at \( x = 1 \), so the absolute minimum is \( f(1) = -7 \).- The largest value is \( 20 \) at \( x = -2 \), so the absolute maximum is \( f(-2) = 20 \).

Key Concepts

Understanding the Derivative

The derivative of a function is a fundamental concept in calculus. It represents the rate at which a function is changing at any point. Think of it as the slope of the tangent line to the function's graph at a particular point.
To find the derivative of a polynomial like the given function, use the power rule. For each term, bring down the exponent as a coefficient, and then subtract one from the exponent. For example, if we have a term like \(2x^3\), the derivative is \(6x^2\).
The derivative helps us identify critical points, which is crucial for finding the maximum and minimum values of the function on an interval.
- It provides a way to find where the function's slope is zero, indicating points where the function might change direction.- This is useful for optimization problems, where we seek to find maximum or minimum values.

Exploring Critical Points

Critical points are values of \(x\) where the derivative of a function is zero or undefined. Finding these points helps us understand where a function might have local maxima or minima.
In the problem above, we found the critical points by setting the derivative \(6x^2 + 6x - 12\) equal to zero, which simplifies to finding where \( (x-1)(x+2) = 0\). This gives us critical points at \(x = -2\) and \(x = 1\).
- By evaluating the function at these critical points, you determine where the function has the potential to turn or change direction.- Not all critical points will be maxima or minima, but they are important locations to consider for optimization.

Finding Absolute Maximum and Minimum

Absolute maxima and minima are the highest and lowest values of a function on a closed interval. To find these, evaluate the function at critical points and endpoints of the interval.
For the function given, we compared values of \(f(x)\) at the endpoints \(x = -3\) and \(x = 2\), and at the critical points \(x = -2\) and \(x = 1\).
- Evaluating functions at these specific \(x\) values gives you different output values, which you then compare.- In our problem, the absolute maximum value was \(f(-2) = 20\), and the absolute minimum value was \(f(1) = -7\).- These evaluations ensure you find the true "tops" and "bottoms"—the highest and lowest peaks of the function on the specified range.