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The function $f(x)=\frac{1}{x+2}$ is one-to-one.


AFind its inverse function $f^{-1}$ and check your answer.
BFind the domain and the range of $f$ and $f^{-1}$.

Verified Answer :

  • Find the inverse \(f^{-1}(x)\) by setting \(y = \frac{1}{x+2}\) and solving for \(x\), resulting in \(f^{-1}(x) = \frac{1-2x}{x} = \frac{1}{x} - 2\).
  • Verify the inverse by checking the compositions \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\).
  • Identify the domain of \(f\) as \(\mathbb{R} \setminus \{-2\}\) and its range as \(\mathbb{R} \setminus \{0\}\).
  • Conclude that the domain of \(f^{-1}\) is \(\mathbb{R} \setminus \{0\}\) and its range is \(\mathbb{R} \setminus \{-2\}\).
  • The final inverse function is \(\boxed{f^{-1}(x) = \frac{1-2x}{x}}\).

Explanation

  1. Analysis of the Function and Goal

We are given the function \(f(x) = \frac{1}{x+2}\), which is defined as a one-to-one mapping. Our goal is to find its inverse function \(f^{-1}(x)\), verify that it is indeed the inverse, and determine the domains and ranges for both \(f(x)\) and \(f^{-1}(x)\). An inverse function \(f^{-1}(y) = x\) exists if the original function \(y = f(x)\) can be solved uniquely for \(x\) in terms of \(y\).

  1. Finding the Inverse Function \(f^{-1}(x)\)

To find the inverse, we set \(y = \frac{1}{x+2}\) and solve for \(x\).
Multiplying both sides by \(x+2\) (given \(x eq -2\)):
\[y(x+2) = 1\]
\[xy + 2y = 1\]
\[xy = 1 - 2y\]
Since \(y eq 0\) (because \(1/(x+2)\) can never be zero), we divide by \(y\):
\[x = \frac{1 - 2y}{y} = \frac{1}{y} - 2\]
Swapping the variables \(x\) and \(y\) to express the function in terms of \(x\), we get:
\[f^{-1}(x) = \frac{1-2x}{x} = \frac{1}{x} - 2\]

  1. Verifying the Inverse Function

We verify the result by checking the compositions \(f(f^{-1}(x))\) and \(f^{-1}(f(x))\):

  1. Compute \(f(f^{-1}(x))\):

\[f(f^{-1}(x)) = f\left(\frac{1}{x} - 2\right) = \frac{1}{(\frac{1}{x} - 2) + 2} = \frac{1}{\frac{1}{x}} = x\]

  1. Compute \(f^{-1}(f(x))\):

\[f^{-1}(f(x)) = f^{-1}\left(\frac{1}{x+2}\right) = \frac{1}{\frac{1}{x+2}} - 2 = (x+2) - 2 = x\]
Since both compositions result in \(x\), the function \(f^{-1}(x) = \frac{1-2x}{x}\) is correctly identified.

  1. Determining Domain and Range

The domain of \(f(x) = \frac{1}{x+2}\) is all real numbers except where the denominator is zero, so \(x eq -2\). The range of \(f(x)\) is all values that \(y = \frac{1}{x+2}\) can take; since the numerator is a non-zero constant, \(y eq 0\).
Using the property of inverse functions:

  • Domain of \(f^{-1}\) = Range of \(f\) = \(\mathbb{R} \setminus \{0\}\).
  • Range of \(f^{-1}\) = Domain of \(f\) = \(\mathbb{R} \setminus \{-2\}\).

This is consistent with the expression \(f^{-1}(x) = \frac{1-2x}{x}\), which is defined for \(x eq 0\), and whose output \(y\) satisfies \(y eq -2\).

Examples

In electronics, the resistance \(R\) of a parallel circuit involving a fixed resistor and a variable resistor can be described by a function similar to \(f(x) = 1/(x+2)\). Finding the inverse function allows an engineer to determine the necessary input variable (like component value) if the desired output resistance is already specified. Understanding these reciprocal relationships is fundamental in designing systems where components influence each other inversely, such as in control theory or circuit analysis.

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