Problem 28 - Calculus Single Variable

Question

Determine the intervals on which the given function \(f\) is concave up, the intervals on which \(f\) is concave down, and the points of inflection of \(f\). Find all critical points. Use the Second Derivative Test to identify the points \(x\) at which \(f(x)\) is a local minimum value and the points at which \(f(x)\) is a local maximum value. $$ f(x)=1 / x+\ln (x) $$

Step-by-Step Solution

Answer

Function is concave up on \((0, 2)\), concave down on \((2, \infty)\), and point of inflection at \(x = 2\). Local minimum at \(x = 1\).

Step 1: Determine the First Derivative

To begin, we need to find the first derivative of the function \(f(x) = \frac{1}{x} + \ln(x)\). Use the power rule and the derivative of a natural logarithm. For \(\frac{1}{x}\), the derivative is \(-\frac{1}{x^2}\). For \(\ln(x)\), the derivative is \(\frac{1}{x}\). Thus, the first derivative \(f'(x)\) is:\[ f'(x) = -\frac{1}{x^2} + \frac{1}{x} \]

Step 2: Simplify the First Derivative

To simplify \(f'(x)\), combine the terms over a common denominator:\[ f'(x) = \frac{-1}{x^2} + \frac{1}{x} = \frac{-1 + x}{x^2} \]

Step 3: Determine Critical Points

Critical points occur where \(f'(x) = 0\) or \(f'(x)\) is undefined. Set the numerator of \(f'(x) = 0\):\[ -1 + x = 0 \]Solving gives \(x = 1\). Since \(f'(x) = \frac{-1 + x}{x^2}\), \(f'(x)\) is undefined at \(x = 0\). Thus, the critical point is \(x = 1\).

Step 4: Determine the Second Derivative

Find the second derivative of \(f(x)\) to analyze concavity and perform the Second Derivative Test. Differentiate \(f'(x) = \frac{-1 + x}{x^2}\):\[ f''(x) = \frac{2 - x}{x^3} \]

Step 5: Analyze Concavity

To find intervals of concavity, determine where \(f''(x)\) is positive or negative. Set \(f''(x) = 0\) and solve:\[ 2 - x = 0 \]This gives \(x = 2\). Analyze the sign of \(f''(x)\) around \(x = 2\):- For \(x < 2\), \(f''(x) > 0\) (concave up).- For \(x > 2\), \(f''(x) < 0\) (concave down).

Step 6: Determine Points of Inflection

A point of inflection occurs where the concavity changes. From Step 5, the concavity changes at \(x = 2\), so \(x = 2\) is a point of inflection.

Step 7: Apply the Second Derivative Test

Use the Second Derivative Test to classify the critical point \(x = 1\). Compute \(f''(1)\): \[ f''(1) = \frac{2 - 1}{1^3} = 1 \] Since \(f''(1) > 0\), \(f\) has a local minimum at \(x = 1\).

Key Concepts

Critical Points

Critical points are significant values of a function that can indicate potential local maxima or minima, or points where the function's behavior changes. To find these points, we need to locate where the first derivative, denoted here as \( f'(x) \), is either zero or undefined.

For the function \( f(x) = \frac{1}{x} + \ln(x) \):

• The first derivative is \( f'(x) = \frac{-1 + x}{x^2} \).
• To find critical points, solve \( f'(x) = 0 \). This happens when the numerator is zero, i.e., \( -1 + x = 0 \). Solving this gives \( x = 1 \).
• Additionally, \( f'(x) \) is undefined at \( x = 0 \), but \( x = 0 \) is outside the domain of \( \ln(x) \) and \( \frac{1}{x} \), so it isn't considered a critical point in this context.

Therefore, our critical point is \( x = 1 \). Critical points help us understand where to apply further tests for local extrema.

Second Derivative Test

The Second Derivative Test is a useful method to classify critical points. This test determines if a critical point is a local maximum, local minimum, or inconclusive.

Here's how it works:

• If \( f''(x) > 0 \) at a critical point \( x \), then \( f(x) \) has a local minimum there.
• If \( f''(x) < 0 \), then \( f(x) \) has a local maximum.
• If \( f''(x) = 0 \), the test is inconclusive.

Apply this to \( x = 1 \) for \( f(x) = \frac{1}{x} + \ln(x) \):

• Calculate \( f''(1) = \frac{2 - 1}{1^3} = 1 \).
• Since \( f''(1) > 0 \), \( x = 1 \) is a point where \( f \) has a local minimum.

The second derivative gives us insights into the behavior and nature of points on the graph of a function.

Concavity Analysis

Concavity tells us how the curve of a function behaves. A function is "concave up" where it resembles a "cup," meaning it curves upwards, and "concave down" where it curves downwards like a hill.

To determine concavity, examine the second derivative \( f''(x) \):

• For "concave up," \( f''(x) > 0 \).
• For "concave down," \( f''(x) < 0 \).

Let's analyze \( f''(x) = \frac{2 - x}{x^3} \):

• Set \( f''(x) = 0 \) to find potential points of inflection: solve \( 2 - x = 0 \), yielding \( x = 2 \).
• Check intervals around \( x = 2 \):
  • For \( x < 2 \), \( f''(x) > 0 \), so \( f \) is concave up.
  • For \( x > 2 \), \( f''(x) < 0 \), so \( f \) is concave down.

By analyzing concavity, we gain insight into the general shape of the graph, helping us predict where and how the function will change.

Points of Inflection

Points of inflection are where the concavity of a function changes from up to down or vice versa. These points show a shift in the curvature of the function's graph.

Finding these points involves:

• Determining where \( f''(x) = 0 \).
• Checking a change in the sign of \( f''(x) \) around these values.

For \( f(x) = \frac{1}{x} + \ln(x) \):

• The second derivative \( f''(x) = \frac{2 - x}{x^3} \) changes sign at \( x = 2 \).

Hence, \( x = 2 \) is a point of inflection for this function.
Understanding points of inflection helps track transitions in the function’s behavior, giving a comprehensive picture of its graph.