Problem 34 - Calculus of a Single Variable

Question

Applying the First Derivative Test In Exercises \(17-40\) , (a) find the critical numbers of \(f\) (if any), (b) find the open interval(s) on which the function is increasing or decreasing, (c) apply the First Derivative Test to identify all relative extrema, and (d) use a graphing utility to confirm your results. $$ f(x)=\frac{x}{x-5} $$

Step-by-Step Solution

Answer

The critical number of the function is \( x = 5 \). The function is decreasing on the intervals \( (-\infty, 5) \) and \( (5, \infty) \). There are no relative extrema.

Step 1: Find the Derivative

The derivative of the function \( f(x)=\frac{x}{x-5} \) can be found by applying the quotient rule. Which is \( f'(a) = \frac{g'(a)h(a) - g(a)h'(a)}{{[h(a)]}^2} \). Therefore, \( f'(x) = \frac{(x-5) - x}{{(x-5)}^2} = \frac{-5}{{(x-5)}^2}

Step 2: Find the Critical Numbers

The critical numbers are where the derivative equals zero or does not exist. Since there are no values for \( x \) which make \( f'(x) = 0 \), the critical number where the derivative is not defined is \( x = 5 \).

Step 3: Determine the Increasing and Decreasing Intervals

This can be done by testing the signs of \( f'(x) \) in the intervals divided by the critical number \( x < 5 \) and \( x > 5 \). For \( x < 5 \), choose \( x = 0 \). Substituting this into \( f'(x) \), you get \( f'(0) = -1 \) which is less than 0 so \( f \) is decreasing on \( (-\infty, 5) \). For \( x > 5 \), choose \( x = 6 \). Substituting this into \( f'(x) \), you get \( f'(6) = -1 \) which is less than 0 so \( f \) is also decreasing on \( (5, \infty) \). Therefore, the function is always decreasing.

Step 4: Apply the First Derivative Test

To apply the First Derivative Test, we need to compare the signs of \( f' \) just to the left and right of each critical number. In this case, \( f' \) changes from negative to negative at \( x = 5 \), so there are no relative extrema.

Step 5: Confirm Results with a Graphing Utility

Using a graphing utility can provide visual confirmation of the previous steps. The function's graph will be a hyperbola, split into two negatively sloped sections alongside the line \( x = 5 \). Hence, there will be no relative maxima or minima, as predicted.

Key Concepts

Critical numbers

Imagine you're on a hike and you're trying to figure out where the trail peaks or dips—that's essentially what we're doing with functions when we look for critical numbers. These are specific spots along the graph of a function where the slope is flat (a derivative of zero), or where the function’s graph takes a turn, but the slope can’t be calculated (where the derivative does not exist).

For the function f(x) = \(\frac{x}{x-5}\), the derivative never equates to zero because \
\(f'(x) = -\frac{5}{{(x-5)}^2}\) is never zero due to the square in the denominator. However, a critical number occurs at \(x=5\) as the function and its derivative aren't defined there. This is a crucial concept for understanding the hills and valleys of a graph—the highs and lows of a function’s journey.

Increasing and decreasing intervals

Let's now explore the map of our mathematical hike to see where the function f(x) = \(\frac{x}{x-5}\) is climbing up or sliding down. This is much like interpreting whether our path goes uphill or downhill.

To do this, after finding our critical number (in this case, \(x=5\)), we check the sign of our derivative, \(f'(x)\), before and after the critical number. If it’s positive, our function is going upwards—increasing. If negative, it's heading downwards—decreasing. In our example, for \(x < 5\) and \(x > 5\), \(f'(x)\) is negative; indicating that the function is consistently decreasing. So, no matter which direction we look from our critical number, our path is always going downhill.

Quotient rule

When dealing with complex fractions in calculus, like when a function is divided by another function, we bring out a handy tool — the quotient rule. It's like a formula that helps us find the slope of a divided path. The quotient rule states that the derivative of a fraction where the numerator is \(g(x)\) and the denominator is \(h(x)\) is given as:

\[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} \]

For our specific function, \(f(x) = \frac{x}{x-5}\), applying the quotient rule allowed us to find its derivative, \(f'(x)\), which led us to understand where the function's graph is going uphill or downhill. This rule is absolutely essential when navigating the ups and downs of rational functions in calculus.

Derivative

In the realms of calculus, the derivative is the wizardly spell that reveals how a function is changing at any given point. It's like having a speedometer that tells you how fast you're going — not in miles per hour, but in terms of how quickly the function’s value is increasing or decreasing.

The derivative of \(f(x)\) at any point 'x' gives you the slope of the tangent line to the function's graph at that point. For our function \(f(x) = \frac{x}{x-5}\), the derivative is \(f'(x) = -\frac{5}{{(x-5)}^2}\), which helps us see the function's behavior. It shows us that no matter where we are on this graph, our function's slope is always negative, meaning our hike is perpetually heading into a decline.