Problem 240 - Calculus Volume 1

Question

Determine a. intervals where \(f\) is increasing or decreasing, b. local minima and maxima of \(f\), c. intervals where \(f\) is concave up and concave down, and d. the inflection points of \(f\). Sketch the curve, then use a calculator to compare your answer. If you cannot determine the exact answer analytically, use a calculator. \(f(x)=\frac{e^{x}}{x}, x \neq 0\)

Step-by-Step Solution

Answer

Increasing for \(x > 1\); decreasing for \(x < 1\). Local min at \(x = 1\). Concave up for \(x > 0\). No inflection points.

Step 1: Find the First Derivative

To determine where the function is increasing or decreasing, we first find the first derivative of the function. Given \( f(x) = \frac{e^x}{x} \), we apply the quotient rule: \( f'(x) = \frac{x(e^x) - e^x(1)}{x^2} = \frac{(x-1)e^x}{x^2} \).

Step 2: Determine Critical Points

Critical points occur where the first derivative is zero or undefined. Set the numerator of \( f'(x) \) to zero: \((x-1)e^x = 0\). This gives the critical point at \( x = 1 \). The derivative is undefined at \( x = 0 \), but since \( x = 0 \) is not in the domain, it isn't considered a critical point.

Step 3: Evaluate Intervals for Increasing or Decreasing

The critical point \( x = 1 \) divides the number line into intervals, \((-fty, 1)\) and \((1, fty)\). Test a point in each interval in \( f'(x) = \frac{(x-1)e^x}{x^2} \). For \( x < 1 \), such as \( x = 0.5 \), \( f'(x) < 0 \) indicating \( f \) is decreasing. For \( x > 1 \), such as \( x = 2 \), \( f'(x) > 0 \) indicating \( f \) is increasing.

Step 4: Identify Local Minima and Maxima

Since \( f(x) \) changes from decreasing to increasing at \( x = 1 \), there is a local minimum at \( x = 1 \).

Step 5: Find the Second Derivative

Find the second derivative to analyze concavity. Differentiate \( f'(x) = \frac{(x-1)e^x}{x^2} \) again using the quotient and product rules to obtain \( f''(x) = \frac{(x^2 + x)e^x - 2x(x-1)e^x}{x^4} \) which simplifies to \( f''(x) = \frac{(x^2 - 2x + 2)e^x}{x^4} \).

Step 6: Evaluate Intervals for Concavity

Set \( f''(x) = 0 \) to find potential inflection points: \( x^2 - 2x + 2 = 0 \) has no real roots as the discriminant is negative, meaning no points where concavity changes calculated through this method. Evaluate \( f''(x) \) for intervals \( (0, fty) \). For \( x > 0 \), say \( x = 1.5 \), \( f''(x) > 0 \); thus, the function is concave up for \( x > 0 \).

Step 7: Sketch and Verify with a Calculator

Sketch the curve by plotting points and using the identified increasing, decreasing, and concave behaviors. Use a graphing calculator to plot the function and verify that the sketch aligns with analytical findings.

Key Concepts

First Derivative

The first derivative of a function, often denoted as \( f'(x) \), provides valuable information about the function's behavior. It tells us whether a function is increasing or decreasing at any given point along its curve. For the function \( f(x) = \frac{e^x}{x} \), the first derivative is found using the quotient rule, yielding \( f'(x) = \frac{(x-1)e^x}{x^2} \).

In this expression, the critical factor impacting whether the function is rising or falling is the sign of \( f'(x) \):

• If \( f'(x) > 0 \), the function is increasing.
• If \( f'(x) < 0 \), the function is decreasing.

Thus, by carefully evaluating \( f'(x) \) over different intervals around the critical points, you can determine where \( f(x) \) is increasing or decreasing.

Critical Points

Critical points in calculus are key to understanding the behavior of a function, especially in identifying local minima or maxima. They occur where the first derivative \( f'(x) = 0 \) or is undefined. For our function \( f(x) = \frac{e^x}{x} \), solving \( (x-1)e^x = 0 \) gives us a critical point at \( x = 1 \).

Critical points serve as indicators of potential changes in the graph:

• At \( x = 1 \), the sign changes in \( f'(x) \) from negative to positive, indicating a local minimum.

Points where the derivative doesn't exist, such as \( x = 0 \) for our function, are not considered since they fall outside the domain we are examining.

Concavity

Concavity refers to the direction the curve bends and influences how the function behaves over intervals. To analyze this, the second derivative \( f''(x) \) is used. For \( f(x) = \frac{e^x}{x} \), we found \( f''(x) = \frac{(x^2 - 2x + 2)e^x}{x^4} \). Evaluating this function over certain intervals helps determine if the curve is concave up or down.

Consider these interpretations:

• If \( f''(x) > 0 \), \( f(x) \) is concave up in that interval.
• If \( f''(x) < 0 \), \( f(x) \) is concave down.

In the given exercise, evaluating \( f''(x) \) shows the curve is concave up for \( x > 0 \). Concavity helps us anticipate the overall shape and form of the graph, crucial for sketching the function.

Inflection Points

Inflection points occur where the function's concavity changes. These points are crucial for understanding the transition in the curve's bending direction. To find them, one would normally solve \( f''(x) = 0 \). However, for the function \( f(x) = \frac{e^x}{x} \), the equation \( x^2 - 2x + 2 = 0 \) results in no real roots. This indicates there are no inflection points derived through this method.

Though inflection points typically mark a change from concave up to concave down (or vice versa), the absence of these points simply tells us the function maintains one kind of concavity across the relevant domain. Doing so solidifies the understanding of the function’s behavior across all analyzed sections.