Problem 40 - Essential Calculus Early Transcendentals

Question

\(37-50=\) Find the absolute maximum and absolute minimum values of \(f\) on the given interval. $$f(x)=x^{3}-6 x^{2}+5, \quad[-3,5]$$

Step-by-Step Solution

Answer

Max value is 5 at \(x = 0\); min value is -76 at \(x = -3\).

Step 1: Understand the function and the interval

We are given the function \( f(x) = x^3 - 6x^2 + 5 \) and we need to find the absolute maximum and minimum values within the interval \([-3,5]\). This involves evaluating the function at critical points and endpoints within the interval.

Step 2: Find the derivative of the function

To find the critical points, first calculate the derivative of \( f(x) \). The derivative \( f'(x) \) is calculated as follows:\[ f'(x) = 3x^2 - 12x \].

Step 3: Solve for critical points

Set the derivative \( f'(x) = 0 \) to find the critical points:\[ 3x^2 - 12x = 0 \].Factor out the common terms:\[ 3x(x - 4) = 0 \].So, the critical points are \( x = 0 \) and \( x = 4 \).

Step 4: Evaluate the function at the critical points and endpoints

Evaluate the function \( f(x) \) at the critical points within the interval and the endpoints:- At \( x = -3 \): \( f(-3) = (-3)^3 - 6(-3)^2 + 5 = -27 - 54 + 5 = -76 \).- At \( x = 0 \): \( f(0) = 0^3 - 6(0)^2 + 5 = 5 \).- At \( x = 4 \): \( f(4) = 4^3 - 6(4)^2 + 5 = 64 - 96 + 5 = -27 \).- At \( x = 5 \): \( f(5) = 5^3 - 6(5)^2 + 5 = 125 - 150 + 5 = -20 \).

Step 5: Identify the absolute maximum and minimum

Compare the values of \( f(x) \) at the endpoints and critical points to determine the absolute extreme values:- The function values:\(-76, 5, -27, -20\).- The absolute maximum value is \( f(0) = 5 \).- The absolute minimum value is \( f(-3) = -76 \).

Key Concepts

Critical Points

Critical points are essential to solving calculus problems, especially when searching for absolute extrema, which is the highest and lowest values a function takes on a specified interval.
To find these points, we look at where the derivative of a function is zero or undefined.
In our exercise, the function is smooth and differentiable everywhere, meaning we only need to find where the derivative equals zero. For the function given, the derivative is set to zero: - Solve the equation \(3x^2 - 12x = 0\).
- You factor it to get \(3x(x - 4) = 0\).
- Thus, the solutions, or critical points, are \(x = 0\) and \(x = 4\). This process shows where the function's slope is zero, indicating possible peaks or troughs, which might signify an extreme point.

Derivative

The derivative of a function is a fundamental concept in calculus. It gives us the rate of change or the slope of the function at any given point. Finding the derivative is the key first step in locating critical points.
For our function, \(f(x) = x^3 - 6x^2 + 5\), the derivative \(f'(x)\) is calculated as follows:

• The derivative of \(x^3\) is \(3x^2\).
• The derivative of \(-6x^2\) is \(-12x\).
• The derivative of a constant, \(5\), is \(0\).

This results in the derivative \(f'(x) = 3x^2 - 12x\).
With this expression, you can determine where the function's rate of change is zero by solving \(f'(x) = 0\), crucial for locating critical points.

Function Evaluation

After finding critical points and considering the derivative, we perform function evaluation.
This step involves plugging the critical points back into the original function to find out their actual values, which help in identifying the extrema. Besides, it is essential to evaluate the function at the endpoints of the interval. For our exercise, the function is evaluated as follows:

• For \(x = -3\), \(f(-3) = -76\).
• For \(x = 0\), \(f(0) = 5\).
• For \(x = 4\), \(f(4) = -27\).
• For \(x = 5\), \(f(5) = -20\).

This comparison of the function's values at its critical points and endpoints will reveal the absolute maximum and minimum values within the interval.

Endpoint Analysis

Endpoints are crucial when determining the absolute extrema of a function over a closed interval. While critical points are found on the interior, absolute extrema can occur at both critical points and endpoints. Evaluating the function at the endpoints helps to ensure no extreme value is overlooked. In our case, the endpoints of the interval \([-3,5]\) are evaluated as follows:

• At \(x = -3\), the function gives \(f(-3) = -76\).
• At \(x = 5\), it results in \(f(5) = -20\).

These values are compared with those at the critical points to find the absolute maximum and minimum. For example, even though critical points might suggest local extrema, the endpoint at \(x = -3\) gives the absolute minimum value in the interval. Therefore, always include endpoints in your evaluation for a complete analysis.