Problem 30 - Calculus Early Transcendentals

Question

Find: (a) the intervals on which f is increasing, (b) the intervals on which f is decreasing, (c) the open intervals on which f is concave up, (d) the open intervals on which f is concave down, and (e) the x-coordinates of all inflection points. \(f(x)=x^{3} \ln x\)

Step-by-Step Solution

Answer

Increasing on \((e^{-\frac{1}{3}}, \infty)\); decreasing on \((0, e^{-\frac{1}{3}})\); concave up on \((e^{-\frac{5}{6}}, \infty)\); concave down on \((0, e^{-\frac{5}{6}})\); inflection point at \(x = e^{-\frac{5}{6}}\).

Step 1: Find the First Derivative

To find increasing and decreasing intervals, we first find the derivative of the function. Start by applying the product rule to \[f(x) = x^3 \ln x.\]The derivative using the product rule \((u'v + uv')\) is:\[f'(x) = (3x^2)(\ln x) + (x^3)(\frac{1}{x}).\]Simplify the expression:\[f'(x) = 3x^2 \ln x + x^2.\]

Step 2: Set First Derivative Equal to Zero

Solve for critical points by setting the first derivative equal to zero:\[3x^2 \ln x + x^2 = 0.\]Factor out \(x^2\):\[x^2(3 \ln x + 1) = 0.\]From this, we have:\[x^2 = 0 \] or \[3 \ln x + 1 = 0.\]For \(x^2 = 0\), \(x = 0\).For \(3 \ln x = -1\), we solve for \(x\) as follows:\[\ln x = -\frac{1}{3},\]\[x = e^{-\frac{1}{3}}.\]

Step 3: Analyze First Derivative

Determine the sign of the first derivative on the intervals around the critical points to find intervals of increase and decrease.Consider the test intervals: - \((0, e^{-\frac{1}{3}})\)- \((e^{-\frac{1}{3}}, \infty)\)Choose test points within these intervals and substitute into \(f'(x) = 3x^2 \ln x + x^2\). - For \(x \in (0, e^{-\frac{1}{3}})\), pick \(x = e^{-1}\): \[f'(e^{-1}) = 3e^{-2} \ln(e^{-1}) + e^{-2} = 3(-e^{-2}) + e^{-2} = -2e^{-2},\] which is negative, so \(f(x)\) is decreasing.- For \(x \in (e^{-\frac{1}{3}}, \infty)\), pick \(x = 1\): \[f'(1) = 3 \cdot 1^2 \ln 1 + 1^2 = 0 + 1 = 1,\] which is positive, so \(f(x)\) is increasing.

Step 4: Find the Second Derivative

To determine concavity, find the second derivative by differentiating \(f'(x)\). Start with:\[f'(x) = 3x^2 \ln x + x^2.\]The derivative is:\[f''(x) = 6x \ln x + 3x + 2x.\]Simplify:\[f''(x) = 6x \ln x + 5x.\]

Step 5: Set Second Derivative Equal to Zero

Find inflection points by setting \(f''(x)\) to zero:\[6x \ln x + 5x = 0.\]Factor out \(x\):\[x(6 \ln x + 5) = 0.\]Thus, \(x = 0\) or \(6 \ln x + 5 = 0\).Solve \(6 \ln x = -5\):\[\ln x = -\frac{5}{6},\]\[x = e^{-\frac{5}{6}}.\]

Step 6: Analyze Second Derivative

Determine concavity by analyzing the second derivative sign on intervals:Consider the test intervals: - \((0, e^{-\frac{5}{6}})\)- \((e^{-\frac{5}{6}}, \infty)\)Choose test points within these intervals and substitute into \(f''(x) = 6x \ln x + 5x\).- For \(x \in (0, e^{-\frac{5}{6}})\), pick \(x = e^{-1}\): \[f''(e^{-1}) = 6e^{-1} \cdot (-1) + 5e^{-1} = -6e^{-1} + 5e^{-1} = -e^{-1},\] which is negative, so \(f(x)\) is concave down.- For \(x \in (e^{-\frac{5}{6}}, \infty)\), pick \(x = 1\): \[f''(1) = 6 \cdot 1 \cdot 0 + 5 \cdot 1 = 0 + 5 = 5,\] which is positive, so \(f(x)\) is concave up.

Key Concepts

Critical Points

Critical points are where the function changes from increasing to decreasing or vice versa. To find the critical points of a function like \( f(x) = x^3 \ln x \), we use its first derivative, as critical points occur where this derivative is zero or undefined.

Applying the product rule, we derive \( f'(x) = 3x^2 \ln x + x^2 \). We then set \( f'(x) \) to zero to investigate possible critical points: \( 3x^2 \ln x + x^2 = 0 \).

By factoring out \( x^2 \), we obtain the equation \( x^2(3 \ln x + 1) = 0 \), leading to critical points at \( x = 0 \) and when \( 3 \ln x + 1 = 0 \).

Solving for \( x \) in \( 3 \ln x = -1 \) gives us \( x = e^{-\frac{1}{3}} \). Critical points are thus \( x = 0 \) and \( x = e^{-\frac{1}{3}} \).

Concavity

Concavity of a function tells us whether it curves upwards or downwards in a specific interval. We determine this by examining the sign of the second derivative.

For \( f(x) = x^3 \ln x \), the second derivative helps us find concavity. We start from \( f'(x) = 3x^2 \ln x + x^2 \) and differentiate again: \( f''(x) = 6x \ln x + 5x \).

We check where \( f''(x) = 6x \ln x + 5x = 0 \) and analyze the sign of \( f''(x) \) in different intervals to determine concavity. If \( f''(x) \) is positive, the graph is concave up. If negative, it's concave down.

Break the intervals into regions around \( x = e^{-\frac{5}{6}} \) to check for sign changes in the second derivative. Calculating \( f''(x) \) indicates intervals of concave up and down.

Inflection Points

Inflection points occur where a function changes its concavity. These are identified using the second derivative, \( f''(x) \).

For \( f(x) = x^3 \ln x \), we solve \( 6x \ln x + 5x = 0 \) to find inflection points.

Factoring gives \( x(6 \ln x + 5) = 0 \), and solving \( 6 \ln x + 5 = 0 \) results in \( x = e^{-\frac{5}{6}} \).

Thus, an inflection point exists at \( x = e^{-\frac{5}{6}} \), where the function transitions between concave up and down.

Derivative Analysis

Derivative analysis is crucial in studying the behavior of functions. It involves taking derivatives to learn about rates of change, slopes, and concavity.

1. **First Derivative**:
- Indicates where the function is increasing or decreasing.
- Critical points occur where the first derivative is zero or undefined.

2. **Second Derivative**:
- Used to find concavity and inflection points.
- If \( f''(x) > 0 \), the function is concave up; if \( f''(x) < 0 \), it's concave down.

Through these derivatives, we derive insights into the function's geometry and help predict behavior across intervals.