Problem 55 - Algebra 2

Question

For each function \(f,\) find \(f^{-1},\) the domain and range of \(f\) and \(f^{-1},\) and determine whether \(f^{-1}\) is a function. $$ f(x)=\frac{1}{(x+1)^{2}} $$

Step-by-Step Solution

Answer

The inverse function is \(f^{-1}(x) = ±\sqrt{\frac{1}{x}} - 1\). The domain of \(f(x)\) is all real numbers except \(x = -1\) and its range is all positive real numbers. The domain of \(f^{-1}(x)\) is all positive real numbers and its range is all real numbers except \(y = -1\). The inverse \(f^{-1}\) is not a function.

Step 1: Find the Inverse Function

Start with the function \(f(x) = \frac{1}{(x+1)^{2}}\). Replace \(f(x)\) with \(y\), this gives us \(y = \frac{1}{(x+1)^{2}}\). Swap \(x\) and \(y\) to get \(x = \frac{1}{(y+1)^{2}}\). Now solve this equation for \(y\). First, rewrite the equation as \((y+1)^{2} = \frac{1}{x}\), then take the square root of both sides to get \(y+1 = ±\sqrt{\frac{1}{x}}\). Finally, subtract 1 from both sides to solve for \(y\), giving us \(y = ±\sqrt{\frac{1}{x}} - 1\). Therefore, the inverse function \(f^{-1}(x) = ±\sqrt{\frac{1}{x}} - 1\).

Step 2: Find the Domains and Ranges

The domain of the initial function \(f(x) = \frac{1}{(x+1)^{2}}\) is all real numbers except \(x = -1\), as this value will make the denominator zero, which is undefined in real numbers. So the domain of \(f(x)\) is \(-∞ < x < -1\) or \(-1 < x < ∞\). The range is all positive real numbers, because the square of any real number is never negative, and since we're taking the reciprocal of it, it will also be positive. So the range is \(0 < y < ∞\).For the inverse function \(f^{-1}(x)\), the domain is all positive real numbers, because we're taking the reciprocal and square root of \(x\), which isn't defined for negative numbers or zero. So the domain is \(0 < x < ∞\). The range is all real numbers except \(y = -1\), due to the subtract 1 operation, so the range is \(-∞ < y < -1\) or \(-1 < y < ∞\).

Step 3: Determine Whether the Inverse is a Function

A relation is considered a function if for every input there is exactly one output. Here, however, the inverse function \(f^{-1}(x) = ±\sqrt{\frac{1}{x}} - 1\) has a plus/minus sign, which gives two possible outputs for each input \(x\). For example, for \(x = 1\), \(y\) could be either \(0\) or \(-2\). Therefore, \(f^{-1}\) is not a function because it doesn't pass the vertical line test.

Key Concepts

Domain and Range

Understanding the concepts of domain and range is essential when dealing with both functions and their inverses. The **domain** of a function is the complete set of possible values of the independent variable, which is usually denoted as x. It tells us what inputs are possible for the function.
Let’s consider the original function in this exercise:

• The domain of \( f(x) = \frac{1}{(x+1)^{2}} \) is all real numbers except \( x = -1 \). At \( x = -1 \), the function becomes undefined due to division by zero.

For the inverse function, things work a little differently:

• The domain of \( f^{-1}(x) = ±\sqrt{\frac{1}{x}} - 1 \) involves only positive real numbers, because square roots and division by zero aren't defined for negative numbers or zero.

Now, the **range** of a function is the complete set of possible values of the dependant variable, typically denoted as y. It tells us what outputs are possible. For the given function:

• The range is all positive real numbers since the output of \( \frac{1}{(x+1)^{2}} \) can never be zero or negative.

In contrast, for the inverse function:

• The range excludes \( y = -1 \) due to the \( -1 \) subtraction in the inverse function, meaning the possible outputs are everything except \(-∞ < y < -1 \) or \(-1 < y < ∞ \).

Function Definition

The definition of a function is fundamental in understanding when a mathematical relation is a function or not. Essentially, a **function** is a relation from a set of inputs to a set of possible outputs where each input is related to exactly one output. In simpler terms, every x from the domain is paired with exactly one y from the range.
For the inverse function in this exercise, the rule \( f^{-1}(x) = ±\sqrt{\frac{1}{x}} - 1 \) provides two potential outputs for any input x, because of the ± symbol.

• This means it fails the core definition of a function as it does not assign a single output to each input.

Let’s consider an example: if \( x = 1 \), then \( y \) can either be \( 0 \) or \( -2 \). Such a situation makes it impossible to call \( f^{-1} \) a true function. It's crucial to remember that for inverse to be a function, each x must connect to only one y, maintaining a one-to-one relationship.

Vertical Line Test

The vertical line test is a simple way to determine if a relation is a function. In this test, imagine drawing vertical lines through a graph of a relation. If each vertical line crosses the graph at no more than one point, the relation is indeed a function.
For the inverse function \( f^{-1}(x) = ±\sqrt{\frac{1}{x}} - 1 \), if we were to plot it, vertical lines would often intersect the graph in more than one place. This directly indicates that:

• This relation has multiple outputs for the same x-value, thus failing the vertical line test.

This failure is due to the \( ± \) symbol, as it introduces two possibilities for y. Visually, it’s easy to spot these failures on a graph if a vertical line crosses at two points.
Remember, a true function must pass the vertical line test. This means reducing any ambiguity in outputs for each input to ensure clean, or singular, intersections with any vertical line.