Problem 30 - 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. $$ f(x)=x-\ln (x)+2 / x $$

Step-by-Step Solution

Answer

The function is concave up for \( x > 0 \), with no points of inflection. A local minimum is at \( x = 2 \).

Step 1: Find the First Derivative

First, we need to calculate the first derivative of the function, \( f(x) = x - \ln(x) + \frac{2}{x} \). The derivative of \( x \) is 1, the derivative of \( -\ln(x) \) is \( -\frac{1}{x} \), and the derivative of \( \frac{2}{x} \) is \( -\frac{2}{x^2} \). So, the first derivative is:\[ f'(x) = 1 - \frac{1}{x} - \frac{2}{x^2} \]

Step 2: Find the Second Derivative

Now, find the second derivative of \( f(x) \) to analyze concavity. Differentiate \( f'(x) = 1 - \frac{1}{x} - \frac{2}{x^2} \):The derivative of \( -\frac{1}{x} \) is \( \frac{1}{x^2} \) and the derivative of \( -\frac{2}{x^2} \) is \( \frac{4}{x^3} \). Therefore, the second derivative is:\[ f''(x) = \frac{1}{x^2} + \frac{4}{x^3} \]

Step 3: Determine Concavity

To determine intervals of concavity, evaluate the sign of \( f''(x) \):- \( f''(x) > 0 \) implies the function is concave up.- \( f''(x) < 0 \) implies the function is concave down.Simplify the expression:\[ f''(x) = \frac{x + 4}{x^3} \]The critical point here is when \( x = -4 \), but since \( x > 0 \) because of the ln function, consider sign changes around \( x = 0 \):- If \( x > 0 \), then \( f''(x) > 0 \). Thus, the function is concave up for \( x > 0 \).

Step 4: Identify Points of Inflection

Points of inflection occur where \( f''(x) = 0 \) or changes sign. However, the only valid concern is when the function transitions through 0, which does not occur in our domain of \( x > 0 \), as \( f''(x) = \frac{x + 4}{x^3} \) remains positive for all \( x > 0 \). Hence, there are no points of inflection for \( x > 0 \).

Step 5: Solve for Critical Points using First Derivative

Set the first derivative \( f'(x) = 1 - \frac{1}{x} - \frac{2}{x^2} = 0 \) to find critical points. Simplify and solve:\[ 1 - \frac{1}{x} - \frac{2}{x^2} = 0 \]Multiply by \( x^2 \):\[ x^2 - x - 2 = 0 \]This factors to \((x - 2)(x + 1) = 0 \), giving solutions \( x = 2 \) and \( x = -1 \) (only \( x = 2 \) is valid for \( x > 0 \)).

Step 6: Apply the Second Derivative Test

Use the second derivative test on \( x = 2 \) to determine local extrema.Since \( f''(2) > 0 \), \( f(x) \) has a local minimum at \( x = 2 \).

Key Concepts

Critical Points

Critical points of a function occur where the first derivative is zero or undefined. These points are crucial in determining potential areas where the function might have local maxima or minima. For the function given, we found the first derivative:

• \( f'(x) = 1 - \frac{1}{x} - \frac{2}{x^2} \)

This needs to be set to zero to find the critical points. In the original solution, the critical equation simplifies to:

• \( 1 - \frac{1}{x} - \frac{2}{x^2} = 0 \)

By solving this, we find that the critical points are \( x = 2 \) and \( x = -1 \). Since the natural logarithm function \( \ln(x) \) is only defined for positive \( x \), only \( x = 2 \) is considered valid in this context. It is always necessary to verify these points further with the Second Derivative Test to classify them as minima or maxima.

Concavity

Understanding concavity is key to getting a grasp on the shape of the graph of a function. A function is

• Concave up when its second derivative \( f''(x) > 0 \).
• Concave down when \( f''(x) < 0 \).

For the function given, we determined:

• \( f''(x) = \frac{1}{x^2} + \frac{4}{x^3} \)

Which simplifies to:

• \( f''(x) = \frac{x + 4}{x^3} \)

The function is analyzed to determine intervals of concavity. In the domain of \( x > 0 \), the expression \( f''(x) \) is always positive. Therefore, the function is concave up for all \( x > 0 \). Concavity provides a visual understanding: concave up implies a U-shaped curve.

Points of Inflection

Points of inflection are where the curve changes concavity, marking a shift from concave up to concave down, or vice versa. For this function, points of inflection occur where the second derivative equals zero or changes sign. Here we solved for:

• \( f''(x) = \frac{x + 4}{x^3} \)

Since \( f''(x) > 0 \) for all \( x > 0 \), this means there are no sign changes, so no points of inflection exist in this region. Defining the domain constraints of our function is crucial, which is implied to be for \( x > 0 \) given the natural log. Therefore, for the domain considered, the function does not exhibit points of inflection, solidifying a consistent concave up nature.

Local Extrema

Local extrema refer to the local minima and maxima within the domain of a function. Once the critical points are identified, the Second Derivative Test becomes a useful tool to determine the nature of these points.

• If \( f''(x) > 0 \) at a critical point, it indicates a local minimum.
• If \( f''(x) < 0 \), it's a local maximum.

Using the critical point \( x = 2 \):

• The second derivative \( f''(2) \) is positive, suggesting that \( x = 2 \) is a point of local minimum.

This test gives a concise determination of where the function climbs to a peak or dips to a trough, refining our understanding of the function's graph.