Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 19-28, determine where the graph of the given function is increasing, decreasing, concave up, and concave down. Then sketch the graph (see Example 4). \(g(x)=x \sqrt{x-2}\)

Step-by-Step Solution

Answer

The function increases and is concave up for \( x > 2 \).

Step 1: Find Critical Points

Firstly, find the first derivative of the function \( g(x) = x \sqrt{x-2} \). Let \( u = x \) and \( v = \sqrt{x-2} \), then, using the product rule, we have:\[g'(x) = \frac{d}{dx}(x) \cdot \sqrt{x-2} + x \cdot \frac{d}{dx}(\sqrt{x-2})\]\[g'(x) = 1 \cdot \sqrt{x-2} + x \cdot \frac{1}{2\sqrt{x-2}}\]Simplifying gives:\[g'(x) = \sqrt{x-2} + \frac{x}{2\sqrt{x-2}}\]Set \( g'(x) = 0 \) to find critical points. The critical points are where the derivative equals zero or is undefined, which occur at \( x = 2 \).

Step 2: Analyze Increasing and Decreasing Intervals

To determine where the function is increasing or decreasing, test the intervals divided by the critical points. Consider the intervals \((2, \infty)\) since \( x \sqrt{x-2} \) is only defined for \(x \ge 2\).- For \( x \in (2, \infty) \), choose a test point like \( x = 3 \): \[g'(3) = \sqrt{3-2} + \frac{3}{2\sqrt{3-2}} = 1 + \frac{3}{2} > 0\]Since \( g'(3) > 0 \), the function is increasing in the interval \( (2, \infty) \).

Step 3: Determine Concavity Using the Second Derivative

Find the second derivative to analyze concavity. Start by differentiating \( g'(x) = \sqrt{x-2} + \frac{x}{2\sqrt{x-2}} \):- \( h(x) = \sqrt{x-2} \), so \( h'(x) = \frac{1}{2\sqrt{x-2}} \)- \( j(x) = \frac{x}{2\sqrt{x-2}} \), using quotient rule:\[j'(x) = \frac{(2\sqrt{x-2})(1) - x(\frac{1}{\sqrt{x-2}})}{(2\sqrt{x-2})^2} = \frac{2(x-2) - x}{4(x-2)} = \frac{x-4}{4(x-2)}\]- Combine to find the second derivative:\[g''(x) = \frac{1}{2\sqrt{x-2}} + \frac{x-4}{4(x-2)}\]Analyze the sign of \( g''(x) \). For concavity \(g'' > 0\), and \(0 < x-2 < 2\) should be checked. Thus, it is crucial to evaluate specifically if it is positive at any specific test points.

Step 4: Determine the Concave Intervals

Analyze for concavity at test points:- Choose \( x = 3 \) (since \(x > 2\)):\[g''(3) = \frac{1}{2\sqrt{3-2}} + \frac{3-4}{4(3-2)} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}\]Since \( g''(3) > 0 \), the function is concave up in \( (2, \infty) \). There are no intervals for \( g''(x) < 0 \) within the domain of the function defined above.

Key Concepts

Critical Points

Critical points are specific values of a function where the derivative is zero or undefined. They are essential in understanding the function's behavior because they can indicate potential maxima, minima, or points of inflection.

To determine the critical points for the function \( g(x) = x \sqrt{x-2} \), we first find the first derivative \( g'(x) \). By using the product rule, we derive:

• \( g'(x) = \sqrt{x-2} + \frac{x}{2\sqrt{x-2}} \)

To find the critical points, set \( g'(x) = 0 \). Here, the derivative is undefined at \( x = 2 \), giving us a critical point.

Thus, checking around this critical point helps us understand the changing nature of the function's slope and identify intervals where the function might change from increasing to decreasing.

Increasing and Decreasing Intervals

Identifying whether a function is increasing or decreasing in particular intervals helps to sketch its overall graph more accurately and understand its behavior better. For \( g(x) = x \sqrt{x-2} \), the derivative \( g'(x) \) helps us determine these intervals.

By testing in the domain, which is \( (2, \infty) \), we select a test point like \( x = 3 \):

• \( g'(3) = 1 + \frac{3}{2} = 2.5 > 0 \)

Because \( g'(3) > 0 \), this indicates that the function is increasing in the interval \( (2, \infty) \).

Through such testing, we understand how the function behaves across its domain and map out where it rises or falls.

Concavity

Concavity concerns how a function curves and involves determining where it is concave up or concave down. This is where the second derivative \( g''(x) \) comes in handy, aiding in deeper insights into the graph’s shape.

The sign of \( g''(x) \) tells us about the concavity:

• \( g''(x) > 0 \) implies the graph is concave up - bowl-like shape.
• \( g''(x) < 0 \) implies the graph is concave down - cap-like shape.

For our function \( g(x) = x \sqrt{x-2} \), computation of \( g''(x) \) yields:

• \( g''(x) = \frac{1}{2\sqrt{x-2}} + \frac{x-4}{4(x-2)} \)

The nature of this derivative gives us insights as we test particular points in the domain.

Second Derivative

The second derivative offers essential information, helping determine not only concavity but also points of inflection where the curve changes its shape.

For \( g(x) = x \sqrt{x-2} \), using differentiation, the second derivative is:

• \( g''(x) = \frac{1}{2\sqrt{x-2}} + \frac{x-4}{4(x-2)} \)

Testing \( x = 3 \) in \( g''(x) \):

• \( g''(3) = \frac{1}{4} > 0 \)

This indicates that the function is concave up as \( g''(3) > 0 \), supporting our earlier findings about the function's behavior in \( (2, \infty) \).

Understanding the second derivative is thus critical in painting a complete picture of the function's behavior and predicting the nature of the graph visually.