Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-10, use the Monotonicity Theorem to find where the given function is increasing and where it is decreasing. \(f(x)=\frac{e^{-x}}{x^{2}}\)

Step-by-Step Solution

Answer

Increasing on \((-\infty, -2)\); decreasing on \((-2, 0) \cup (0, \infty)\).

Step 1: Differentiate the Function

First, find the derivative of the function \( f(x) = \frac{e^{-x}}{x^2} \) by using the quotient rule. The quotient rule is \( \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \), where \( u = e^{-x} \) and \( v = x^2 \). Compute the derivative: \[ f'(x) = \frac{-(e^{-x})(x^2) - 2x(e^{-x})}{x^4} = \frac{-e^{-x}(x^2 + 2x)}{x^4}.\]

Step 2: Simplify the Derivative

Factor out common terms from the numerator of the derivative \( f'(x) = \frac{-e^{-x}(x^2 + 2x)}{x^4} \). This becomes \[ f'(x) = \frac{-e^{-x}x(x + 2)}{x^4} = \frac{-e^{-x}(x + 2)}{x^3}. \]

Step 3: Determine Critical Points

Find the critical points by setting the derivative equal to zero or undefined. The function is undefined at \( x = 0 \), and setting \( -e^{-x}(x + 2) = 0 \) gives \( x + 2 = 0 \), so \( x = -2 \). Thus, the critical point is \( x = -2 \).

Step 4: Test Intervals for Monotonicity

Divide the x-axis into intervals based on the critical points: \((-\infty, -2)\), \((-2, 0)\), and \((0, \infty)\). Test points from each interval in the derivative \( f'(x) = \frac{-e^{-x}(x + 2)}{x^3} \) to determine the sign of \( f'(x) \). For \( x = -3 \), \( f'(-3) > 0 \), meaning increasing on \((-\infty, -2)\). For \( x = -1 \), \( f'(-1) < 0 \), meaning decreasing on \((-2, 0)\). For \( x = 1 \), \( f'(1) < 0 \), meaning decreasing on \((0, \infty)\).

Step 5: State Conclusion

From the test intervals, we determine that \( f(x) \) is increasing on \((-\infty, -2)\) and decreasing on \((-2, 0) \cup (0, \infty)\).

Key Concepts

Quotient Rule

Differentiating functions that are written as a ratio of two expressions can be managed by using the quotient rule. This rule is a fundamental tool in calculus, especially when dealing with problems involving division. The rule requires you to identify two functions, often denoted as \( u \) and \( v \), where both are differentiable. For the function \( f(x) = \frac{e^{-x}}{x^2} \), we apply the quotient rule: \( \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \). Here, \( u = e^{-x} \) and \( v = x^2 \).

When applying this formula, it's important to find \( u' \), the derivative of the numerator \( u \), which is \( -e^{-x} \), and \( v' \), the derivative of \( v \), which is \( 2x \). The quotient rule combines these derivatives to effectively find the derivative of the entire function. This process can feel complex but breaking it down into parts simplifies it.

Derivative

The derivative provides crucial information about the function's rate of change and behavior. In our function \( f(x) = \frac{e^{-x}}{x^2} \), the derivative helps us understand how the function is changing at any given point. After applying the quotient rule, we find the derivative: \[ f'(x) = \frac{-e^{-x}(x^2 + 2x)}{x^4} \]. Simplifying this gives \[ f'(x) = \frac{-e^{-x}(x + 2)}{x^3} \].

This simplification is crucial for analyzing the function's behavior more easily. It's notable that as the numerator of \( f'(x) \) changes sign, it suggests that the original function may be increasing or decreasing—key information that will help us determine critical points and intervals of monotonicity.

Critical Points

Critical points are where the derivative of a function is zero or undefined. These points are essential in identifying where the function changes direction, from increasing to decreasing or vice versa. For the function \( f(x) = \frac{e^{-x}}{x^2} \), we set the derivative \( f'(x) = \frac{-e^{-x}(x + 2)}{x^3} \) equal to zero or undefined to find these points.

In this case, the derivative is undefined at \( x = 0 \) due to division by zero. Setting the numerator to zero, \( -e^{-x}(x + 2) = 0 \), leads to \( x + 2 = 0 \) or \( x = -2 \). Thus, we have critical points at \( x = 0 \) and \( x = -2 \). These points help divide the number line into intervals which we can analyze to see the behavior of the function.

Increasing and Decreasing Functions

Analyzing whether a function is increasing or decreasing on certain intervals helps us understand the function's overall shape and behavior. We use the derivative to find these intervals. After finding the critical points and simplifying the derivative \( f'(x) = \frac{-e^{-x}(x + 2)}{x^3} \), we test intervals around these points.

Let's consider intervals \((-fty, -2)\), \((-2, 0)\), and \((0, \infty)\). By testing a point from each interval, you check the sign of \( f'(x) \).

• For \( x = -3 \), \( f'(-3) > 0 \): the function is increasing on \((-fty, -2)\).
• For \( x = -1 \), \( f'(-1) < 0 \): the function is decreasing on \((-2, 0)\).
• For \( x = 1 \), \( f'(1) < 0 \): the function is decreasing on \((0, \infty)\).

Thus, the function increases and decreases on these intervals, helping to visualize its overall trend.