Question

The function \(f\) is one-to-one. (a) Find its inverse function \(f^{-1}\) and check your answer. (b) Find the domain and the range of \(f\) and \(f^{-1}\). $$ f(x)=-\frac{3 x+1}{x} $$

Short answer

The inverse function is \( f^{-1}(x) = -\frac{1}{x + 3} \). The domain of \( f \) is \( \mathbb{R} \setminus \{0\} \), and its range is \( \mathbb{R} \setminus \{-3\} \). The domain of \( f^{-1} \) is \( \mathbb{R} \setminus \{-3\} \), and its range is \( \mathbb{R} \setminus \{0\} \).

Step-by-step solution

- Understand the given function

The given function is \( f(x) = -\frac{3x + 1}{x} \). It's a rational function, meaning its values depend on the division of two polynomials.

- Find the inverse function

To find the inverse function, start by replacing \( f(x) \) with \( y \): \[ y = -\frac{3x + 1}{x} \]Next, swap \( x \) and \( y \) to find the inverse: \[ x = -\frac{3y + 1}{y} \]Now solve for \( y \) to get the inverse function. Multiply both sides by \( y \):\[ xy = -(3y + 1) \] Which simplifies to \[ xy + 3y = -1 \]Factor out \( y \):\[ y(x + 3) = -1 \]Divide both sides by \( x + 3 \):\[ y = -\frac{1}{x + 3} \]Thus, the inverse function is\( f^{-1}(x) = -\frac{1}{x + 3} \).

- Verify the inverse function

To check the inverse function, compose \( f \) with \( f^{-1} \): \[ f(f^{-1}(x)) = f\left(-\frac{1}{x + 3}\right) \]Substitute \( f(x) \) with the inverse:\[ f\left(-\frac{1}{x + 3}\right) = -\frac{3\left(-\frac{1}{x + 3}\right) + 1}{-\frac{1}{x + 3}} \]Simplify inside the fraction:\[ -\frac{-\frac{3}{x + 3} + 1}{-\frac{1}{x + 3}} = -\frac{\frac{1 \cdot (x + 3) - 3}{x + 3}}{-\frac{1}{x + 3}} \]Resulting in:\[ -\frac{x}{1} = x \]Thus, \( f(f^{-1}(x)) = x \). Similarly, verify \( f^{-1}(f(x)) \) and ensure it also results in \(x\) to confirm that the found inverse \( f^{-1} \) is correct.

- Find the domain and range of \( f \)

To find the domain of \( f(x) \), set the denominator not equal to zero: \[ x eq 0 \] Thus, the domain of \( f \) is all real numbers except \( x = 0 \), i.e., \( \text{Domain of } f = \mathbb{R} \setminus \{0\} \).To find the range, observe that \( f(x) \=-\frac{3x + 1}{x} \) can take any real value except the value that would make the expression undefined.

- Find the domain and range of \( f^{-1} \)

For the domain of \( f^{-1}(x) = -\frac{1}{x + 3} \), set the denominator not equal to zero: \[ x + 3 eq 0 \rights] x eq -3 \] Hence, the domain of \( f^{-1} \) is all real numbers except \( x = -3 \), i.e., \( \text{Domain of } f^{-1} = \mathbb{R} \setminus \{-3\} \).The range of \( f^{-1} \) is the same as the domain of \( f \), i.e., \( \text{Range of } f^{-1} = \mathbb{R} \setminus \{0\} \).

Key concepts

One-to-One Functions

A one-to-one function (or injective function) is a type of function where each value of the output variable corresponds to exactly one value of the input variable. This means no two different inputs can produce the same output. Ensuring that a function is one-to-one is crucial for determining its inverse.

For instance, the function given in the original exercise is a one-to-one function. This property allows us to find its inverse by swapping the input and output variables and then solving for the new output. It's essential to note that only one-to-one functions have inverses that are also functions.

One way to check if a function is one-to-one is by using the horizontal line test on its graph. If no horizontal line intersects the graph of the function more than once, the function is one-to-one.

Rational Functions

Rational functions are functions that can be expressed as the quotient of two polynomials, where the denominator is not equal to zero. The function given in the exercise is a rational function:

The analysis of rational functions involves understanding how the numerator and denominator influence the behavior of the function.

When dealing with rational functions, it is also important to consider any values that make the denominator zero, as these cause the function to be undefined. In our example, the function is undefined at 0, which is a critical point in determining the domain.

Additionally, rational functions often have vertical asymptotes at points where the denominator is zero and horizontal asymptotes that describe the behavior of the function as the input grows very large in positive or negative directions. Understanding these characteristics helps in graphing the function and finding its domain and range.

Domain and Range

The domain of a function includes all possible input values (x-values) that the function can accept without resulting in an undefined expression. For the function given, , the domain is all real numbers except those that make the denominator zero. This means our domain is all real numbers except 0.

The range of a function consists of all possible output values (y-values) the function can produce. For rational functions, the range can often be evaluated by considering the values the function approaches as the input becomes large or small. For the given function, you must reason out the range by examining the function's behavior.

For the inverse function, its domain is determined similarly, i.e., by ensuring the denominator does not make the function undefined. Since the inverse function is also rational, its domain would exclude values that lead to zero denominators. The range of the inverse function is the same as the domain of the original function.