Problem 23 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, find the absolute extrema of the function on the closed interval. Use a graphing utility to verify your results. $$ f(x)=x^{3}-3 x^{2}, \quad[-1,3] $$

Step-by-Step Solution

Answer

The absolute minimum of the function \(f(x)=x^{3}-3x^{2}\) on the interval \([-1,3]\) is \(-4\) at \(x=2\). There is no absolute maximum.

Step 1: Find the Derivative

Start by taking the derivative \(f'(x)\) of the given function \(f(x)=x^{3}-3x^{2}\) to find the critical points. So, \(f'(x)=3x^{2}-6x\).

Step 2: Find the Critical Points

Next, set the derivative equal to zero to find the critical points: \(3x^{2}-6x=0\). Factoring the equation yields \(x(3x-6)=0\), resulting to \(x=0\) and \(x=2\) as potential critical points.

Step 3: Evaluating the function

Evaluate the function at the critical points and the end points of the interval \([-1,3]\). Hence the points are \(-1,0,2,3\). So, \(f(-1)=-2\), \(f(0)=0\), \(f(2)=-4\), and \(f(3)=0\).

Step 4: Determine the absolute extrema

Having computed values in the previous step, we can determine that the minimum value is \(-4\) and occurs at \(x=2\). There is no absolute maximum on the interval as \(f(x)\) approaches \(\infty\) as \(x\) approaches \(\infty\) and \(-\infty\) as \(x\) approaches \(-\infty\).

Key Concepts

Critical Points

In calculus, critical points refer to the values of x at which the derivative of a function either does not exist or is equal to zero. These points are crucial because they help us analyze the function's behavior and are potential locations of relative maxima and minima. To find the critical points of a function, like our exercise with the function f(x) = x^3 - 3x^2, you need to:

• Take the derivative of the function, f'(x).
• Set this derivative equal to zero and solve for x.
• Also consider points where the derivative does not exist.

In our case, after setting the derivative f'(x) = 3x^2 - 6x equal to zero, we factored it to find x = 0 and x = 2 as our critical points.

Derivative of Function

The derivative of a function represents the rate at which the function's value is changing at any given point. For a function f(x), the derivative is typically denoted as f'(x). When we take the derivative of a polynomial function, like f(x) = x^3 - 3x^2, we apply power rules, which involve bringing down the exponent as a multiplier and reducing the exponent by one. With polynomials, it's straightforward:

• For x^3, the derivative is 3x^2.
• For -3x^2, the derivative is -6x.

This gives us the derivative, f'(x) = 3x^2 - 6x, which we then use to find critical points and assess the function's behavior.

Graphing Utility Verification

Using a graphing utility provides a visual representation of the function and its behavior on the given interval. After identifying critical points and evaluating the function's value at those points, a graphing utility allows us to visually verify the results. It shows us where the function increases or decreases, and where it attains its highest and lowest values within the interval. In the exercise, we can graph the cubic function f(x) = x^3 - 3x^2 over the interval [-1, 3] and should expect to see a local minimum at x = 2, corresponding to the function's absolute minimum value within the interval, as the analysis suggests.

Function Evaluation

When we speak of function evaluation, we mean calculating the value of a function f(x) at particular points x. This process is fundamental in finding the absolute extrema of a function on a closed interval. To evaluate a function:

• Substitute the x value into the function.
• Perform the indicated operations.

For instance, to evaluate our function f(x) = x^3 - 3x^2 at x = -1, we calculate f(-1) = (-1)^3 - 3(-1)^2 = -1 - 3 = -4. Evaluation at critical points and at the endpoints of the interval, as in our exercise, reveals where the function attains its extreme values on that interval. We discovered that f(2) = -4 is the minimum, and concluded by also evaluating at the interval's endpoints.