Problem 21 - BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

Question

Determine whether the inverse of \(f\) is a function. Then find the inverse. \(f(x)=\frac{1}{2 x-1}\)

Step-by-Step Solution

Answer

Yes, the inverse of the function is itself a function. The inverse function is \(f^{-1}(x) = \frac{1 + x}{2x}\).

Step 1: Verify the function is invertible

The given function \(f(x) = \frac{1}{2 x - 1}\) is one-to-one as it passes the horizontal line test. That means for every \(y\) in the range of the function, there exists a unique \(x\) in the domain. Therefore, this function is invertible.

Step 2: Find the inverse function

To find the inverse, first replace \(f(x)\) with \(y\), so we have: \(y = \frac{1}{2x - 1}\). Then, switch the roles of \(y\) and \(x\) to get: \(x = \frac{1}{2y - 1}\). This is the inverse relation, and we need to solve this for \(y\) to get the inverse function. Multiply both sides by \((2y-1)\) to clear the fraction, resulting in: \(x(2y - 1) = 1\), then distribute the \(x\), so you get: \(2xy - x = 1\). Isolate \(y\) on one side of the equation. First add \(x\) to both sides, to obtain: \(2xy = 1 + x\). Then divide each side by \(2x\), to find: \(y = \frac{1 + x}{2x}\). So the inverse function is \(f^{-1}(x) = \frac{1 + x}{2x}\).

Step 3: Verify the inverse is a function

The inverse \(f^{-1}(x) = \frac{1 + x}{2x}\) passes the vertical line test. This means for every \(x\) in the domain of the inverse function, there corresponds a unique \(y\) in its range. Therefore, the inverse is a function itself.

Key Concepts

Horizontal Line Test

The horizontal line test is a useful method in determining whether a function is one-to-one, which is crucial when deciding if an inverse function exists. A function is one-to-one if no horizontal line intersects the graph of the function more than once. This ensures that each output (\(y\)) is produced by exactly one input (\(x\)).
In the context of the given function\(f(x) = \frac{1}{2x - 1}\), applying this test helps verify that the function can be inverted. If the function passes the horizontal line test, as is the case here, it confirms that for every output, there's a distinct input, indicating the function's invertibility.

Vertical Line Test

The vertical line test is a different tool that's applied to verify if a relation or a set of points is actually a function. It states that if a vertical line intersects the graph of a relation more than once, then the relation is not a function.
This test is vital when examining inverses. In our exercise, once the inverse function\(f^{-1}(x) = \frac{1 + x}{2x}\)was found, it was important to ensure it was a function. Applying the vertical line test verifies that for every \(x\) in the domain, there is only one \(y\) in the range, affirming that the inverse relation is indeed a function.

One-to-One Function

A one-to-one function ensures that each \(x\)-value in the domain is paired with a unique \(y\)-value in the range. This property is key when considering inverse functions. If a function is not one-to-one, it lacks an inverse that is itself a function.

• For example, the function \(f(x) = \frac{1}{2x - 1}\) is one-to-one because it passes the horizontal line test revealing distinct \(x\)-values for each \(y\)-value.
• This one-to-one nature of the function guarantees the existence of an inverse that maps uniquely from its range back to its domain.

Domain and Range

Understanding the domain and range of a function and its inverse is fundamental. The domain of a function is the complete set of possible \(x\)-values, whereas the range is the complete set of possible \(y\)-values. When finding inverses, these roles are swapped.
For the function \(f(x) = \frac{1}{2x - 1}\), the domain excludes any \(x\)-value that makes the denominator zero, hence, domain is \(x eq 0.5\). The range is any real number since the fraction \(\frac{1}{2x - 1}\) can approach any finite value.
When inverted, the original range becomes the domain of the inverse function, \(f^{-1}(x) = \frac{1 + x}{2x}\), and the domain becomes the range, highlighting the deep interplay between these two sets.