Problem 33 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, use a graphing utility to find graphically the absolute extrema of the function on the closed interval. $$ f(x)=0.4 x^{3}-1.8 x^{2}+x-3, \quad[0,5] $$

Step-by-Step Solution

Answer

The absolute maximum of the function in the interval is 11.5 at \( x=5 \), and the absolute minimum is -4.05 at \( x=1.5 \)

Step 1: Find the derivative of the function

The derivative of the function \( f(x)=0.4 x^{3}-1.8 x^{2}+x-3 \) function is: \( f'(x)=1.2x^{2}-3.6x+1 \)

Step 2: Find where the derivative is zero

The derivative \( f'(x)=1.2x^{2}-3.6x+1 \) equals zero where \( x=1.5 \) and \( x=0.55 \). These are the critical points inside the interval and must be checked for extrema.

Step 3: Evaluate the function at endpoints and critical points

Evaluate the function \( f(x) \) at the critical points and endpoints of the interval. So we check \( f(0), f(0.55), f(1.5), f(5) \). We find that \( f(0)=-3, f(0.55)= -2.6, f(1.5)= -4.05, f(5)= 11.5 \).

Step 4: Identify absolute extrema

Compare the values obtained in the previous step to identify the absolute extrema. The absolute maximum is 11.5 at \( x=5 \), and the absolute minimum is -4.05 at \( x=1.5 \)

Key Concepts

Derivative

Before diving into extrema, let's explore the concept of a **derivative**. A derivative represents the rate at which a function is changing at any given point. In simpler terms, it's like having a speedometer for a function. It tells us how fast the function's value is climbing or decreasing.
The derivative of a function, often denoted as \( f'(x) \), is critical when finding extrema. Why? Because it helps pinpoint where changes occur from increasing to decreasing, or vice versa. This precise location is where potential maxima and minima lie.
For the function \( f(x)=0.4x^{3}-1.8x^{2}+x-3 \), its derivative is \( f'(x)=1.2x^{2}-3.6x+1 \). Calculating derivatives might need some practice, but once you conquer this, you're on the path to mastering function behavior analysis!

Critical Points

Now that we understand derivatives, let's address **critical points**. These points are where the derivative equals zero or doesn't exist. Essentially, they're the spots where a function's slope is flat, indicating a possible peak (maximum) or valley (minimum).
To find critical points in the example function, set the derivative \( f'(x)=1.2x^{2}-3.6x+1 \) equal to zero and solve for \( x \). Here, the critical points are \( x=1.5 \) and \( x=0.55 \).
Recognizing critical points is crucial, as they indicate where to look for potential extrema. Though not every critical point is an extremum, examining these points is an essential step in your calculus toolkit.

Evaluating Endpoints

Alongside critical points, it's crucial to evaluate **endpoints** of the interval, as extrema can lurk here too. Endpoints are the boundaries of our interval — in this exercise, that's \([0, 5]\).
When we evaluate the function at endpoints, we simply plug them into the original function \( f(x) \). So, check \( f(0) \) and \( f(5) \).
This method ensures we don't overlook any absolute extrema, as sometimes maximum or minimum values occur at these boundary points rather than only within the interior of the interval.

Graphing Utility

Finally, let's turn our attention to **graphing utilities**. These are modern marvels that allow us to visualize functions as graphs, making the hunt for extrema much simpler.
Tools like graphing calculators or software can plot the function over the specified interval, like \([0, 5]\) for our function. With these graphs, peaks and troughs become visually apparent.
Utilizing a graphing utility is an excellent way to double-check your calculus work, ensuring accuracy in finding extrema. It provides a visual confirmation that complements the analytical process of derivatives and critical points.