Problem 17 - The Calculus with Analytic Geometry

Question

Find the absolute extrema of the given function on the given interval, if there are any, and find the values of \(x\) at which the absolute extrema occur. Draw a sketch of the graph of the function on the interval. $$ f(x)=\frac{4}{(x-3)^{2}},[2,5] $$

Step-by-Step Solution

Answer

Absolute maximum value is 4 at x = 2. Absolute minimum value is 1 at x = 5.

Step 1 - Identify the function and interval

The given function is \( f(x) = \frac{4}{(x-3)^2} \) and the interval is \([2, 5]\).

Step 2 - Find the critical points

To find the critical points, take the derivative of the function and set it to zero. The derivative of \( f(x) \) is: \[ f'(x) = -\frac{8}{(x-3)^3} \] Since this derivative is never zero, there are no critical points within the interval.

Step 3 - Evaluate the function at the endpoints

Calculate \( f(x) \) at the endpoints of the interval \[ f(2) = \frac{4}{(2-3)^2} = \frac{4}{1} = 4 \] \[ f(5) = \frac{4}{(5-3)^2} = \frac{4}{4} = 1 \]

Step 4 - Determine the absolute extrema

Since there are no critical points, compare the values at the endpoints: \( f(2) = 4 \) and \( f(5) = 1 \). The absolute maximum is \( f(2) = 4 \) and the absolute minimum is \( f(5) = 1 \).

Step 5 - Sketch the graph

On the interval \([2, 5]\), the function \( f(x) = \frac{4}{(x-3)^2} \) has a vertical asymptote at \( x = 3 \). It decreases as \( x \) moves away from 3 in either direction. At \( x = 2 \), the function value is 4, and at \( x = 5 \), the function value is 1. Sketch these points and the overall shape of the graph, showing the asymptotic behavior at \( x = 3 \).

Key Concepts

Critical Points in Calculus

Critical points play an essential role in finding the absolute extrema of a function. To locate these points, differentiate the function and set the derivative equal to zero.
Next, solve for the values of \( x \). However, in some cases, like the function \( f(x) = \frac{4}{(x-3)^2} \), the derivative may not have any zero points.
For example, the derivative of our function is: \[ f'(x) = -\frac{8}{(x-3)^3} \]
Since the derivative never equals zero, there are no critical points.
Remember, to confirm if a critical point is an extremum, you would typically check the second derivative or use the first-derivative test.
But in situations like this, where no critical points exist in the provided interval, you will need to evaluate endpoints and analyze the overall behavior of the function.

Evaluating Functions at Endpoints

When a function has no critical points within an interval, determining the absolute extrema requires evaluating the function at the interval’s endpoints.
In our example, we evaluate the given function \( f(x) = \frac{4}{(x-3)^2} \) at \( x = 2 \) and \( x = 5 \).
Here is how you can evaluate the function at the endpoints:

• Calculate \( f(2) = \frac{4}{(2-3)^2} = 4 \)
• Calculate \( f(5) = \frac{4}{(5-3)^2} = 1 \)

By comparing these values, we can determine:

• The absolute maximum is \( f(2) = 4 \)
• The absolute minimum is \( f(5) = 1 \)

This method is particularly useful when dealing with functions that do not have critical points within the specified interval.

Vertical Asymptotes in Functions

A vertical asymptote refers to a line \( x = a \) where a function's value grows infinitely as it approaches \( a \).
For instance, in the function \( f(x) = \frac{4}{(x-3)^2} \), there’s a vertical asymptote at \( x = 3 \).
This is because the denominator approaches zero as \( x \rightarrow 3 \), causing \( f(x) \) to grow without bound.
When sketching the function, it's crucial to illustrate this asymptotic behavior.
Here’s how to consider asymptotes:

• Identify the vertical asymptote by determining where the denominator goes to zero.
• Analyze the behavior of the function as it approaches the asymptote from both sides.
• In our example, as \( x \) moves away from 3, the function values decrease in both directions.

Remember to include the function values at key points along the interval when sketching.
For example, at \( x = 2 \), \( f(x) = 4 \), and at \( x = 5 \), \( f(x) = 1 \).
These values, combined with the asymptotic behavior, will help you create an accurate sketch of the function's graph.