Problem 40 - College Algebra

Question

For the following exercises, find the inverse of the function and graph both the function and its inverse. $$ f(x)=\frac{1}{x^{2}}, x \geq 0 $$

Step-by-Step Solution

Answer

The inverse function is \( f^{-1}(x) = \frac{1}{\sqrt{x}} \), for \( x > 0 \).

Step 1: Understanding the Function

We have the function \( f(x) = \frac{1}{x^2} \) with the condition \( x \geq 0 \). This means the function is defined for non-negative values of \( x \).

Step 2: Setting Up for Inverse Function

To find the inverse, we start by replacing \( f(x) \, \text{with} \, y \). This gives us \( y = \frac{1}{x^2} \). Our job is to solve for \( x \) in terms of \( y \).

Step 3: Solving for x in Terms of y

Rearrange the equation to express \( x \) in terms of \( y \): \( y = \frac{1}{x^2} \) can be rewritten as \( x^2 = \frac{1}{y} \). Solving for \( x \), we get \( x = \sqrt{\frac{1}{y}} \), and since \( x \geq 0 \), \( x = \frac{1}{\sqrt{y}} \). Thus, the inverse function is \( f^{-1}(y) = \frac{1}{\sqrt{y}} \) for \( y > 0 \).

Step 4: State the Inverse Function

The inverse function is \( f^{-1}(x) = \frac{1}{\sqrt{x}} \), defined for \( x > 0 \).

Step 5: Graphing the Function and Its Inverse

Graph the original function \( y = \frac{1}{x^2} \) and its inverse \( y = \frac{1}{\sqrt{x}} \). The graphs will reflect across the line \( y = x \).

Key Concepts

Function Transformation

Function transformation involves modifying a function to achieve a desired form. For the function given in the exercise, which is \( f(x) = \frac{1}{x^2} \) for \( x \geq 0 \), transformation plays a key role when finding the inverse function. When working with transformations, keep these points in mind:

• The inverse transformation exchanges the roles of \( x \) and \( y \), essentially flipping the function across the line \( y = x \).
• This specific transformation involves algebraic manipulation, which directly leads to finding the inverse.
• Apply transformations carefully to correctly hold onto any restrictions, such as \( x \geq 0 \) in this exercise.

The process of switching \( x \) and \( y \) reflects the core principle of finding inverse functions, directly impacting the transformation applied to the function.

Graphing Functions

Graphing functions makes abstract mathematical concepts more tangible. When graphing \( f(x) = \frac{1}{x^2} \), it is important to remember:

• The function is only defined for \( x \geq 0 \).
• As \( x \) increases, \( f(x) \) decreases, approaching zero, but never reaching it.
• The graph is a downward opening parabola that is steepest near the origin on the positive side.

Graphing its inverse \( f^{-1}(x) = \frac{1}{\sqrt{x}} \) involves similar steps and helps visualize the relationship between a function and its inverse.

To graph the inverse:

• The inverse is defined for \( x > 0 \) and decreases as \( x \) increases.
• Plotting both graphs on the same axes illustrates their reflection symmetry across the line \( y = x \).
• Symmetry confirms the correctness of inverse calculations and provides visual insight into their relationships.

Domain and Range

Understanding the domain and range of a function is essential for graphing both the function and its inverse. For the function \( f(x) = \frac{1}{x^2} \):

• Domain: Since \( x \geq 0 \), the domain is \( [0, \infty) \).
• Range: As \( f(x) \) \( \geq 0 \) for all \( x \), the range is \((0, \infty) \).

For the inverse function \( f^{-1}(x) = \frac{1}{\sqrt{x}} \):

• Domain: Since \( x > 0 \), the domain is \( (0, \infty) \).
• Range: The range aligns with the domain of the original function: \( [0, \infty) \).

Understanding these relationships between the original function and its inverse ensures appropriately defined graphs on a coordinate plane.

Calculus

Although the task at hand primarily involves algebraic transformations and graphing, calculus often plays a supporting role. In this context, calculus helps us:

• Analyze the behavior of both the function and its inverse, particularly their limits as they approach endpoints of their domains.
• Confirm that the transformations applied hold true across all required intervals.
• Calculate derivative functions, which can further deepen understanding of the function's behavior, although not explicitly covered here.

Using calculus can improve intuition and verification when working through the relationships and transformations necessary with inverse functions.