Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

The function is decreasing for all \( x > 0 \).

Step 1: Understanding the Monotonicity Theorem

The Monotonicity Theorem states that if the derivative of a function, \( f'(x) \), is positive on an interval, then \( f(x) \) is increasing on that interval. Conversely, if \( f'(x) \) is negative, then \( f(x) \) is decreasing.

Step 2: Find the Derivative of the Function

Find the derivative \( f'(x) \) of the function \( f(x) = \frac{e^{-x}}{x^2} \) using the quotient rule. The quotient rule states \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \), where \( u = e^{-x} \) and \( v = x^2 \).

Step 3: Apply the Quotient Rule

Compute \( u' = -e^{-x} \), \( v = x^2 \), \( v' = 2x \).Using the quotient rule:\[ f'(x) = \frac{(-e^{-x})(x^2) - (e^{-x})(2x)}{(x^2)^2} = \frac{-e^{-x}x^2 - 2e^{-x}x}{x^4}\]

Step 4: Simplify the Derivative

Factor out \( -e^{-x}x \) from the numerator:\[ f'(x) = \frac{-e^{-x}x(x + 2)}{x^4} = \frac{-e^{-x}(x + 2)}{x^3} \].

Step 5: Determine Where the Derivative is Positive or Negative

The derivative \( f'(x) = -\frac{e^{-x}(x + 2)}{x^3} \) is negative whenever \( x \) is positive because the exponent and the polynomial terms always keep the signs consistent for positive \( x \). Also, it's worth noting the critical points and the function behavior as \( x \) approaches zero, keeping the domain in mind as \( x = 0 \) and negatives would imply undefined parts (such as division by zero or complex values in real scenario).

Step 6: Conclusion on Increase and Decrease

Since \( f'(x) = -\frac{e^{-x}(x + 2)}{x^3} \) is negative for all \( x > 0 \), the function \( f(x) = \frac{e^{-x}}{x^2} \) is decreasing on its entire domain of \( x > 0 \).

Key Concepts

Derivative

The derivative of a function is a fundamental concept in calculus. It measures how a function changes as its input changes. In other words, it tells us the rate at which the function's value is being altered at any given point. For a function \(f(x)\), the derivative is represented as \(f'(x)\). The derivative provides critical information about the function's behavior, such as where it increases or decreases. To find the derivative, different rules are applied depending on the structure of the function. For example, the quotient rule (which we will discuss next) is essential for functions that are ratios of two expressions.

Quotient Rule

The quotient rule is a technique used to find the derivative of a quotient of two functions. When you have a function \(f(x)\) expressed as \(u(x)/v(x)\), where both \(u\) and \(v\) are functions of \(x\), the quotient rule states that the derivative \(\left(\frac{u}{v}\right)'\) can be found using the formula:

• \( f'(x) = \frac{u'v - uv'}{v^2} \)

Here, \(u'\) is the derivative of \(u(x)\), and \(v'\) is the derivative of \(v(x)\). This rule is particularly useful when dealing with ratios, like in our given function \(f(x) = \frac{e^{-x}}{x^2} \). By applying the quotient rule, we can decompose the function into the derivatives of its numerator and denominator, helping us understand the function's rate of change effectively. This method simplifies finding derivatives by systematically breaking down parts of the function.

Increasing and Decreasing Functions

A crucial part of understanding functions is determining where they increase or decrease. This analysis involves looking at the derivative of the function. If the derivative \(f'(x)\) is positive over an interval, the function is said to be increasing on that interval. If \(f'(x)\) is negative, the function is decreasing. These intervals are where the function's output, or y-values, rise or fall as \(x\) increases. Knowing these intervals helps mathematicians and students understand the behavior of the function across its domain. In our problem, after finding \(f'(x)\), we examine where it is positive or negative to determine the increasing or decreasing nature of the function \(f(x) = \frac{e^{-x}}{x^2} \). According to the exercise, the derivative remains negative for all \(x > 0\), which indicates that the function is decreasing in this entire interval.

Critical Points

Critical points of a function occur where its derivative is zero or undefined. These points are significant because they often indicate potential extremes or changes in the function's behavior. To identify critical points, one must first find the derivative and solve for values of \(x\) that make the derivative either zero or undefined. These critical points can then be further analyzed using tests like the first or second derivative test to determine if they are locations of maximum, minimum, or points of inflection. In the given exercise, the derivative \(f'(x) = -\frac{e^{-x}(x + 2)}{x^3}\) is undefined at \(x=0\) and negative for all \(x > 0\), indicating no extrema but complex behavior near \(x = 0\) due to division by zero. Understanding where the derivative is undefined helps clarify the function's overall behavior and guides proper domain consideration.