Problem 8 - Essential Calculus Early Transcendentals

Question

(a) Find the intervals on which \(f\) is increasing or decreasing. (b) Find the local maximum and minimum values of \(f .\) (c) Find the intervals of concavity and the inflection points. \(f(x)=x^{2} \ln x\)

Step-by-Step Solution

Answer

(a) Decreasing on \((0, e^{-1/2})\), increasing on \((e^{-1/2}, \infty)\). (b) Local min at \(x = e^{-1/2}\), value \(-\frac{1}{2e}\). (c) Concave down on \((0, e^{-3/2})\), concave up on \((e^{-3/2}, \infty)\), inflection at \(x = e^{-3/2}\).

Step 1: Find the first derivative

To find the intervals where the function is increasing or decreasing, we first find the derivative of the function. Given the function \(f(x) = x^2 \ln x\), we can use the product rule:\[f'(x) = \frac{d}{dx}(x^2 \ln x) = \frac{d}{dx}(x^2) \cdot \ln x + x^2 \cdot \frac{d}{dx}(\ln x)\]This gives us:\[f'(x) = 2x \ln x + x^2 \cdot \frac{1}{x} = 2x \ln x + x\]

Step 2: Find critical points

We set the derivative equal to zero to find critical points:\[f'(x) = 2x \ln x + x = 0\]Factor out \(x\):\[x(2 \ln x + 1) = 0\]This gives critical points at \(x = 0\) (which we cannot use since \(\ln x\) is undefined there) and when \(2 \ln x + 1 = 0\). Solving \(2 \ln x + 1 = 0\) gives \(\ln x = -\frac{1}{2}\), thus \(x = e^{-1/2}\).

Step 3: Test intervals for increasing or decreasing

Test intervals around the critical point \(x = e^{-1/2}\). Choose test points in intervals \((0, e^{-1/2})\) and \((e^{-1/2}, \infty)\).For the interval \((0, e^{-1/2})\), choose \(x = e^{-1}\):\[f'(e^{-1}) = 2(e^{-1}) \ln(e^{-1}) + e^{-1} = -2e^{-1} + e^{-1} = -e^{-1} < 0\].For the interval \((e^{-1/2}, \infty)\), choose \(x = 1\):\[f'(1) = 2(1) \ln(1) + 1 = 0 + 1 = 1 > 0\].Function \(f(x)\) is decreasing on \((0, e^{-1/2})\) and increasing on \((e^{-1/2}, \infty)\).

Step 4: Identify local extrema

From the above, the function changes from decreasing to increasing at \(x = e^{-1/2}\), indicating a local minimum.Evaluate \(f(x)\) at \(x = e^{-1/2}\):\[f(e^{-1/2}) = (e^{-1/2})^2 \ln(e^{-1/2}) = e^{-1} \times -\frac{1}{2} = -\frac{1}{2e}\].There is a local minimum at \(x = e^{-1/2}\) with value \(-\frac{1}{2e}\).

Step 5: Find the second derivative for concavity

For concavity, calculate the second derivative: \[f''(x) = \frac{d}{dx}(2x \ln x + x) = \frac{d}{dx}(2x \ln x) + \frac{d}{dx}(x)\]Using the product rule on \(2x \ln x\):\[\frac{d}{dx}(2x \ln x) = 2\ln x + 2 \cdot \frac{d}{dx}(x \ln x) = 2\ln x + 2\]Thus, \[f''(x) = \frac{d}{dx}(2x \ln x) + \frac{d}{dx}(x) = 2\ln x + 2 + 1/x\]\[= 2 \ln x + 3\].

Step 6: Determine intervals of concavity and inflection points

Set \(f''(x) = 0\) for possible inflection points:\[2\ln x + 3 = 0\]\[\ln x = -\frac{3}{2}\]\[x = e^{-3/2}\]Test intervals around \(x = e^{-3/2}\) to determine concavity:- For \((0, e^{-3/2})\), choose \(x = e^{-2}\):\[f''(e^{-2}) = 2\ln(e^{-2}) + 3 = -4 + 3 = -1 < 0\] (concave down).- For \((e^{-3/2}, \infty)\), choose \(x = 1\):\[f''(1) = 2 \ln(1) + 3 = 3 > 0\] (concave up).The function is concave down on \((0, e^{-3/2})\) and concave up on \((e^{-3/2}, \infty)\). There is an inflection point at \(x = e^{-3/2}\).

Key Concepts

Increasing and Decreasing Intervals

To determine where a function is increasing or decreasing, we first need to find its derivative. This derivative helps identify the rate of change of the function. For the function \( f(x) = x^2 \ln x \), we use the product rule to find its derivative:

• Product Rule Formula: \((uv)' = u'v + uv'\)
• Result: \( f'(x) = 2x \ln x + x \)

Critical points occur where the derivative equals zero or is undefined. In this example, solving \( 2x \ln x + x = 0 \) yielded the critical point \( x = e^{-1/2} \). Testing around this point helps us conclude:

• Decreasing on the interval \((0, e^{-1/2})\)
• Increasing on the interval \((e^{-1/2}, \infty)\)

Recognizing increasing or decreasing intervals is key to understanding the behavior of functions over different domains.

Concavity

Concavity describes the direction a function curves, which is dictated by the second derivative. For our function \( f(x) = x^2 \ln x \), after finding the first derivative, we compute the second derivative:

• First derivative: \( f'(x) = 2x \ln x + x \)
• Second derivative: \( f''(x) = 2 \ln x + 3 \)

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

• If \( f''(x) > 0 \), the function is concave up (curves upwards).
• If \( f''(x) < 0 \), the function is concave down (curves downwards).

For example, testing intervals:

• Concave down on \((0, e^{-3/2})\)
• Concave up on \((e^{-3/2}, \infty)\)

Understanding concavity is crucial to predict the behavior of a graph's shape and identify turning points.

Inflection Points

An inflection point is where a function changes its concavity, transitioning from concave up to concave down or vice versa. Such points could influence the overall trend or shape of a graph significantly.From our function example, we found:

• The second derivative: \( f''(x) = 2 \ln x + 3 \)
• Setting this to zero: \( \ln x = -\frac{3}{2} \), gives \( x = e^{-3/2} \)

This solution indicates an inflection point at \( x = e^{-3/2} \). Changes in sign around this point from the concave down to concave up confirm an inflection takes place here.Identifying inflection points helps in analyzing changes in the "bending" of a graph and can assist in making more nuanced predictions about the function’s behavior.

Local Maxima and Minima

Local maxima and minima are points where the function reaches a highest or lowest value in its vicinity. Finding these involves using the first derivative to locate critical points, then testing around them to see the change in behavior of the function.For \( f(x) = x^2 \ln x \):

• Critical point found at \( x = e^{-1/2} \)
• Sign change from \( f'(x) < 0 \) to \( f'(x) > 0 \) indicates a local minimum
• Value at this point: \( f(e^{-1/2}) = -\frac{1}{2e} \)

Detecting local maxima or minima helps in identifying peaks and valleys within a function, which can be crucial in diverse applications like optimizing processes or interpreting physical phenomena.