Problem 73 - Precalculus Student Solutions Manual 5th

Question

For each function as defined that is one-to-one, (a) write an equation for the inverse function in the form y = ƒ -11x2, (b) graph ƒ and ƒ -1 on the same axes, and (c) give the domain and the range of ƒ and ƒ -1. If the function is not one-to-one, say so. $$f(x)=\frac{-3 x+12}{x-6}, \quad x \neq 6$$

Step-by-Step Solution

Answer

(a) \( f^{-1}(x) = \frac{12+6x}{x+3} \), (b) Graph the function and its inverse symmetrically along \( y=x \), (c) Domain of \( f \) is \( x eq 6 \), range is \( y eq -3 \); Domain of \( f^{-1} \) is \( x eq -3 \), range is \( y eq 6 \).

Step 1: Verify One-to-One Function

Check if the function \( f(x) = \frac{-3x+12}{x-6} \) is one-to-one by ensuring it passes the Horizontal Line Test.

Step 2: Write the Equation for the Inverse Function

Start by setting \( y = \frac{-3x+12}{x-6} \). Then solve for \( x \) to express it in terms of \( y \). Swap \( x \) and \( y \) at the end to find the inverse.

Step 3: Algebraic Manipulation

Multiply both sides by \( x-6 \): \[ y(x-6) = -3x + 12 \]Distribute and rearrange terms to isolate \( x \): \[ yx - 6y = -3x + 12 \]Move all \( x \)-terms to one side: \[ yx + 3x = 12 + 6y \]Factor out \( x \): \[ x(y + 3) = 12 + 6y \]Solve for \( x \): \[ x = \frac{12 + 6y}{y + 3} \]Thus, the inverse function is: \[ f^{-1}(x) = \frac{12 + 6x}{x + 3} \]

Step 4: Graph the Function and its Inverse

Graph \( f(x) = \frac{-3x+12}{x-6} \) and \( f^{-1}(x) = \frac{12+6x}{x+3} \) on the same axes. The graphs should be symmetrical along the line \( y = x \).

Step 5: Determine the Domain and Range

For \( f(x) = \frac{-3x+12}{x-6} \), the domain is \( x eq 6 \) and the range is \( y eq -3 \). For \( f^{-1}(x) = \frac{12+6x}{x+3} \), the domain is \( x eq -3 \) and the range is \( y eq 6 \).

Key Concepts

One-to-One Functions

A function is one-to-one if every output value corresponds to exactly one input value. This means different inputs produce different outputs. Checking if a function is one-to-one is crucial before finding its inverse.

To determine if a function is one-to-one, you can use the horizontal line test.

If no horizontal line intersects the graph of the function more than once, the function is one-to-one.

For the function \( f(x) = \frac{-3x+12}{x-6} \), we need to check if it passes the horizontal line test.

Horizontal Line Test

The horizontal line test is a visual way to determine if a function is one-to-one. Imagine drawing horizontal lines all over the graph of the function:

- If any horizontal line crosses the function more than once, it's not a one-to-one function.
- If no horizontal line crosses more than once, the function is one-to-one.

This test works because it ensures each output is associated with a single input, which is the essence of a one-to-one function.

For our function \( f(x) = \frac{-3x+12}{x-6} \), we can graph it to verify:

- The function \( f(x) = \frac{-3x+12}{x-6} \) passes the horizontal line test, so it is one-to-one.

Now we can find its inverse.

Domain and Range

When discussing functions, two important concepts are domain and range.

The domain of a function is the set of all possible input values, while the range is the set of all possible output values.

For the function \( f(x) = \frac{-3x+12}{x-6} \):

• The domain excludes the value that makes the denominator zero, so the domain is \( x eq 6 \).


• The range excludes the value that the function cannot output, so the range is \( y eq -3 \).

For its inverse \( f^{-1}(x) = \frac{12 + 6x}{x + 3} \):

• The domain excludes the value that makes its denominator zero, so the domain is \( x eq -3 \).


• The range excludes the value the function cannot output, so the range is \( y eq 6 \).