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 \(f^{-1}\). $$ f(x)=\frac{2 x+3}{x+2} $$

Short answer

The inverse function is \( \frac{3 - 2x}{x - 2} \). The domain of \( f \) is \( x eq -2 \) and its range is \( y eq 2 \). The domain of \( f^{-1} \) is \( x eq 2 \) and its range is \( y eq -2 \).

Step-by-step solution

- Express the Function

Given the function: \[ f(x)=\frac{2x+3}{x+2} \]

- Replace f(x) with y

Start by replacing \(f(x)\) with \(y\): \[ y = \frac{2x+3}{x+2} \]

- Swap x and y

Next, swap \(x\) and \(y\): \[ x = \frac{2y+3}{y+2} \]

- Solve for y

Solve for \(y\) by isolating it on one side of the equation: \[ x(y+2) = 2y + 3 \] \[ xy + 2x = 2y + 3 \] \[ xy - 2y = 3 - 2x \] \[ y(x-2) = 3 - 2x \] \[ y = \frac{3 - 2x}{x - 2} \]

- Write the Inverse Function

The inverse function is: \[ f^{-1}(x) = \frac{3 - 2x}{x - 2} \]

- Verify the Inverse Function

To verify, check that \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \): \[ f(f^{-1}(x)) = f\left( \frac{3 - 2x}{x - 2} \right) = x \] \[ f^{-1}(f(x)) = f^{-1}\left( \frac{2x+3}{x+2} \right) = x \]. Both conditions hold.

- Find the Domain and Range of f(x)

For \( f(x) = \frac{2x+3}{x+2} \), the domain is all real numbers except where the denominator is zero: \[ x eq -2 \]. The range is all real numbers except the value that makes the numerator and denominator equal: \[ y = 2 \]

- Find the Domain and Range of f^{-1}(x)

For \( f^{-1}(x) = \frac{3 - 2x}{x - 2} \), the domain is all real numbers except where the denominator is zero: \[ x eq 2 \]. The range is all real numbers except the value that makes the numerator and denominator equal: \[ y = -2 \]

Key concepts

Domain and Range

The domain of a function is the set of all possible input values (x-values) that the function can accept. For the function \(f(x)=\frac{2x+3}{x+2}\) The expression The domain of \(f(x)\) includes all real numbers except where the denominator is zero, since a denominator of zero is undefined: \(x eq -2\). Hence, the domain of \(f(x)\) is all real numbers except \(-2\).To find the range of \(f(x)\), we look at the set of possible output values (y-values). We take into account where the function cannot take a value. By analyzing the expression \(f(x)=\frac{2x+3}{x+2}\), we see that the function cannot output 2 because this value would require us to divide by zero. Thus, the range of \(f(x)\) is all real numbers except \(2\).For the inverse function \(f^{-1}(x) = \frac{3 - 2x}{x - 2}\), the domain is all real numbers except \(2\) (where the denominator does not make sense), and the range is all real numbers except \(-2\).

One-to-One Function

A one-to-one function is a function where each output value corresponds exactly to one input value, and no input value maps to the same output value. This property is essential for a function to have an inverse, since every output must map back uniquely to one input.For our function \(f(x)=\frac{2x+3}{x+2}\) , we need to verify that it is one-to-one. To check this, we use the horizontal line test. If any horizontal line crosses the graph of the function at most once, the function is one-to-one.Additionally, ensuring that the derivative is always positive or always negative can also help confirm one-to-one-ness. f'(x) = \frac{2(x+2)- (2x+3)}{(x+2)^2} Checking the expressions, we can see that our function passes these tests, proving it's one-to-one.

Algebraic Manipulation

To find the inverse of a function algebraically, we need several steps of algebraic manipulation. For the function \(f(x)=\frac{2x+3}{x+2}\):

• Replace f(x) with y: \(y = \frac{2x+3}{x+2}\)
• Swap x and y: \(x = \frac{2y+3}{y+2}\)
• Solve for y:
\begin{itemize}
\( x(y+2) = 2y + 3 \)
\(xy + 2x = 2y + 3 \)
\(xy - 2y = 3 - 2x \)
\( y(x-2) = 3 - 2x \)

\(y = \frac{3 - 2x}{x - 2}\)

After these steps, we have the inverse function \( f^{-1}(x) = \frac{3 - 2x}{x - 2}\).Verifying the inverse involves checking if \( f(f^{-1}(x)) = x \) and \(f^{-1}(f(x)) = x\). Working systematically through algebraic manipulation helps ensure our solutions are correct.