Problem 238 - Calculus Volume 1

Question

For the following exercises, determine $$ \begin{array}{l}{\text { a. intervals where } f \text { is increasing or decreasing, }} \\ {\text { b. local minima and maxima of } f} \\ {\text { c. intervals where } f \text { is concave up and concave }} \\ {\text { down, and }} \\ {\text { d. the inflection points of } f \text { . Sketch the curve, then }} \\ {\text { use a calculator to compare your answer. If you }} \\\ {\text { cannot determine the exact answer analytically, use }} \\ {\text { a calculator. }}\end{array} $$ $$ f(x)=\ln x \sqrt{x}, x>0 $$

Step-by-Step Solution

Answer

\(f(x)\) is decreasing on \((0, e^{-2})\) and increasing on \((e^{-2}, \infty)\) with a local minimum at \(x = e^{-2}\). It is concave up on \((0, 1)\) and concave down on \((1, \infty)\), with an inflection point at \(x = 1\).

Step 1: Find the First Derivative

To determine the intervals where the function is increasing or decreasing, we need to find the first derivative of the function. Given the function \( f(x) = \ln x \cdot \sqrt{x} \), we can use the product rule to find its derivative. The product rule states \((uv)' = u'v + uv'\). Let \( u = \ln x \) and \( v = \sqrt{x} = x^{1/2} \). Then:\[ u' = \frac{1}{x}, \quad v' = \frac{1}{2}x^{-1/2} \]The first derivative is:\[ f'(x) = \left(\ln x\right) \cdot \frac{1}{2}x^{-1/2} + x^{1/2} \cdot \frac{1}{x} \]Which simplifies to:\[ f'(x) = \frac{\ln x}{2 \sqrt{x}} + \frac{1}{\sqrt{x}} = \frac{\ln x + 2}{2 \sqrt{x}} \]

Step 2: Determine Increasing/Decreasing Intervals

Examine the sign of \( f'(x) = \frac{\ln x + 2}{2 \sqrt{x}} \). The critical point occurs where \( f'(x) = 0 \):\[ \ln x + 2 = 0 \]\[ \ln x = -2 \]\[ x = e^{-2} \]Check sign changes around the critical point (\(x = e^{-2}\)) by evaluating \(f'(x)\) for intervals. For \(x < e^{-2}\), choose \(x = 0.1\): \(f'(0.1) < 0\). For \(x > e^{-2}\), choose \(x = 1\): \(f'(1) > 0\). Thus, \(f(x)\) is decreasing in \((0, e^{-2})\) and increasing in \((e^{-2}, \infty)\).

Step 3: Identify Local Minima and Maxima

Since \(f(x)\) changes from decreasing to increasing at \(x = e^{-2}\), there is a local minimum at \(x = e^{-2}\). Substitute back into \(f(x)\) to find the location: \(f(e^{-2}) = \ln(e^{-2}) \cdot \sqrt{e^{-2}} = -2 \cdot e^{-1} = -\frac{2}{e}\).

Step 4: Find the Second Derivative

To examine concavity and find inflection points, compute the second derivative \(f''(x)\). Starting from:\[ f'(x) = \frac{\ln x + 2}{2 \sqrt{x}} \]Using the quotient rule \((\frac{u}{v})' = \frac{u'v - uv'}{v^2}\), we find:\[ u = \ln x + 2, \quad v = 2\sqrt{x} \]\[ u' = \frac{1}{x}, \quad v' = \frac{1}{\sqrt{x}} \]\[ f''(x) = \frac{(\frac{1}{x})(2\sqrt{x}) - (\ln x + 2)(\frac{1}{\sqrt{x}})}{4x} \]Simplifying, we've:\[ f''(x) = \frac{2 - (\ln x + 2)}{4x\sqrt{x}} \]\[ f''(x) = \frac{2 - \ln x - 2}{4x\sqrt{x}} \]\[ f''(x) = \frac{-\ln x}{4x\sqrt{x}} \]

Step 5: Determine Concavity Intervals

To find where \(f(x)\) is concave up or down, consider \(f''(x) = \frac{-\ln x}{4x\sqrt{x}} \). The inflection point is where \(f''(x) = 0\):\[ -\ln x = 0 \]\[ \ln x = 0 \]\[ x = 1 \]For concavity intervals, check \(f''(x)\) around \(x = 1\). For \(x < 1\), choose \(x = 0.5\): \(f''(0.5) > 0\), concave up. For \(x > 1\), choose \(x = 2\): \(f''(2) < 0\), concave down. Thus, \(f(x)\) is concave up for \((0, 1)\) and concave down for \((1, \infty)\).

Step 6: Identify Inflection Points

Since the concavity changes at \(x = 1\), there is an inflection point at \(x = 1\). Calculate \(f(1) = \ln(1) \cdot \sqrt{1} = 0\). The inflection point is \((1, 0)\).

Step 7: Sketch and Verify with Calculator

Plot the function based on intervals of increase, decrease, and concavity. Verify key features such as local extrema and inflection points with a graphing calculator. This confirms the analytic findings.

Key Concepts

Derivative

In calculus, the derivative of a function represents the rate at which the function's value changes as its input changes. Essentially, it describes the function's slope at any point. For the exercise, we focused on finding the first derivative of the function \( f(x) = \ln x \sqrt{x} \). By applying the product rule, which states \((uv)' = u'v + uv'\), the first derivative \( f'(x) \) was determined to be \( \frac{\ln x + 2}{2 \sqrt{x}} \). This derivative helps us find where the function is increasing or decreasing, giving us insight into the behavior of the function over its domain.

Concavity

Concavity describes the curvature of a graph. A function is concave up when its graph looks like a right-side-up bowl, and concave down when it resembles an upside-down bowl. To find concavity, we need the second derivative of the function. For \( f(x) = \ln x \sqrt{x} \), we calculated the second derivative \( f''(x) = \frac{-\ln x}{4x\sqrt{x}} \). By examining the sign of this second derivative, we can tell where the curve is concave up or down. In this exercise, we found:

• Concave up on \((0, 1)\)
• Concave down on \((1, \infty)\)

These intervals explain the shape of the function's graph across its domain.

Inflection Points

Inflection points occur where a function changes concavity. This means the graph of the function shifts from concave up to concave down or vice versa. To find inflection points, we set the second derivative equal to zero and solve for \(x\). For this function, solving \( f''(x) = \frac{-\ln x}{4x\sqrt{x}} = 0 \) leads to an inflection point at \( x = 1 \). At this point, the function value is \( f(1) = 0 \), confirming the inflection point is \((1, 0)\). This marks a notable change in the shape of the graph.

Local Minima

In calculus, a local minimum is where a function reaches a lower value than all nearby points, forming a valley-like structure. We find local minima by examining where the first derivative changes signs and solving it for zero. In this exercise, \( f'(x) = \frac{\ln x + 2}{2 \sqrt{x}} = 0 \) gave us the critical point \( x = e^{-2} \). Since the derivative changes from negative to positive here, we confirmed a local minimum exists. We calculated \( f(e^{-2}) = -\frac{2}{e} \), making \( (e^{-2}, -\frac{2}{e}) \) the coordinates of the local minimum.

Local Maxima

A local maximum is a point where a function value is higher than all other nearby values, creating a peak. For this function \( f(x) \), we examined the critical points to determine if any were local maxima. By checking the derivative signs before and after each critical point, we determine if there's a transition from positive to negative, which indicates a local maximum. However, for this particular exercise, no such transition occurs, meaning the function has no local maxima within the domain considered. Each critical point was evaluated, ensuring accuracy in identifying such features.