Problem 26 - Thomas Calculus

Question

In Exercises \(9-28 :\) a. Find the intervals on which the function is increasing and decreasing. b. Then identify the function's local extreme values, if any, saying where they are taken on. c. Which, if any, of the extreme values are absolute? d. Support your findings with a graphing calculator or computer grapher. $$ g(x)=x^{2 / 3}(x+5) $$

Step-by-Step Solution

Answer

The function is decreasing on \((-\infty, -8]\) and increasing on \((-8, \infty)\). It has a local minimum at \(x = -8\), which is not absolute.

Step 1: Finding the First Derivative

To determine where the function is increasing or decreasing, we first need to find its derivative. The function is given by \[ g(x) = x^{2/3}(x + 5). \]Using the product rule \[ (uv)' = u'v + uv', \]let \( u = x^{2/3} \) and \( v = x + 5 \). Then, \( u' = \frac{2}{3}x^{-1/3} \) and \( v' = 1 \). Applying the product rule:\[ g'(x) = \frac{2}{3} x^{-1/3}(x + 5) + x^{2/3}. \]

Step 2: Simplifying the Derivative

Simplify \( g'(x) \) to find the critical points:\[g'(x) = \frac{2}{3} x^{-1/3}(x + 5) + x^{2/3}= \frac{2}{3} x^{2/3} + \frac{10}{3} x^{-1/3} + x^{2/3}= \frac{5}{3} x^{2/3} + \frac{10}{3} x^{-1/3}. \]

Step 3: Finding Critical Points

To find critical points, set \( g'(x) = 0 \) and solve:\[\frac{5}{3} x^{2/3} + \frac{10}{3} x^{-1/3} = 0, \]which simplifies to:\[5x^{2/3} + 10x^{-1/3} = 0 \Rightarrow 5x^{2/3} = -10x^{-1/3}.\]Simplifying further gives \( x = -8 \) as a critical point. Also consider points where \( g'(x) \) is undefined, including when \( x = 0 \).

Step 4: Analyzing Increasing/Decreasing Intervals

Determine the sign of \( g'(x) \) around the critical points. Test intervals around \( x = -8 \) and \( x = 0 \) using test point values in \( g'(x) \):- For \( x < -8 \), choose \( x = -27 \).- For \( -8 < x < 0 \), choose \( x = -1 \).- For \( x > 0 \), choose \( x = 1 \).Evaluate \( g'(x) \) for these test points to determine sign, indicating increasing/decreasing intervals.

Step 5: Identifying Local Extreme Values

Use the results of increasing and decreasing intervals to identify local extrema:- A change from negative to positive in \( g'(x) \) at \( x = -8 \) indicates a local minimum.- Since \( g'(x) \) does not change sign around \( x = 0 \), no extrema exist there. Thus, the local minimum is at \( x = -8 \).

Step 6: Determining Absolute Extreme Values

Compare the local minimum \( g(-8) \) with the behavior of \( g(x) \) as \( x \to \pm \infty \) to check for absolute extrema. Since the function only decreases and increases around \( x = -8 \) and goes to infinity as \( x \to \pm \infty \), there are no absolute extrema.

Step 7: Graphical Support

Use a graphing tool to plot \( g(x) = x^{2/3}(x + 5) \). Confirm that the graph increases and decreases consistent with our earlier analysis, showing a local minimum at \( x = -8 \) and no absolute extrema over the real numbers.

Key Concepts

First Derivative

Understanding the first derivative is crucial in calculus as it helps us determine the behavior of a function. For a given function, the first derivative can provide information about how the function is changing, or its rate of change, at any point on its curve. This rate of change indicates whether the function is increasing or decreasing.

In our exercise, the function is given by \[ g(x) = x^{2/3}(x + 5) \]. To find the first derivative, we use the product rule, which is \[ (uv)' = u'v + uv' \] for functions \(u\) and \(v\). Here, \( u = x^{2/3} \) and \( v = x + 5 \). We differentiate these and find\[ u' = \frac{2}{3} x^{-1/3} \text{ and } v' = 1 \]. Applying the product rule gives us:\[ g'(x) = \frac{2}{3}x^{-1/3}(x + 5) + x^{2/3} \].

The resulting expression tells us how the function changes over its domain, and this derivative will help locate critical points, and hence determine increasing and decreasing intervals.

Critical Points

Critical points of a function occur where its first derivative is either zero or undefined. These points are essential because they provide insights into the function's behavior and can indicate potential locations for local extrema.

In this example, to find the critical points, we set the first derivative equal to zero:\[ \frac{5}{3} x^{2/3} + \frac{10}{3} x^{-1/3} = 0 \]. Solving this equation gives us the critical point \( x = -8 \). Additionally, we must consider where the derivative is undefined, such as \( x = 0 \) due to the term \( x^{-1/3} \) becoming undefined. Listing these points helps us further investigate the function's nature and distinguish regions of different behavior in the function's graph.

Increasing and Decreasing Intervals

Once we have the critical points, we can use them to find where the function is increasing or decreasing. Increasing intervals occur where the first derivative is positive, and decreasing intervals where it is negative.

To determine this, we take a small test point from each interval around our critical points \( x = -8 \text{ and } x = 0 \). For example, choosing \( x = -27 \) for the interval \( x < -8 \), \( x = -1 \) for the interval \( -8 < x < 0 \), and \( x = 1 \) for \( x > 0 \) will allow us to check the sign of \( g'(x) \).

• If \( g'(x) > 0 \), the function is increasing in that interval.
• If \( g'(x) < 0 \), the function is decreasing.

Through this process, we learn that the function decreases for \( x < -8 \), increases for \( -8 < x < 0 \), and again increases for \( x > 0 \).

Local Extrema

Local extrema refer to the points in the function where it reaches a local maximum or minimum value. These usually occur at critical points where the derivative changes sign.

For this function, at \( x = -8 \), \( g'(x) \) changes from negative to positive, indicating a local minimum. Since the derivative around \( x = 0 \) does not change sign, we do not have any extrema there.

Local extrema are important in understanding the shape of the graph and how the function behaves. In this scenario, the presence of a local minimum shows a dip in the graph at that point. However, as the function tends to increase indefinitely as \( x \) moves away from this minimum point, there are no absolute extrema. Utilizing a graphing calculator can visualize this behavior, confirming the local minimum at \( x = -8 \) and the lack of absolute maximum or minimum values on the entire real line.