Problem 65 - A Graphical Approach to College Algebra

Question

For each function that is one-to-one, write an equation for the inverse function in the form \(y=f^{-1}(x)\) and then graph \(f\) and \(f^{-1}\) on the same axes. Give the domain and range of \(f\) and \(f^{-1}\). If the function is not one-to-one, say so. $$f(x)=\sqrt{6+x}, x \geq-6$$

Step-by-Step Solution

Answer

The inverse is \( f^{-1}(x) = x^2 - 6 \); both are reflections over \( y=x \).

Step 1: Determine if the Function is One-to-One

A function is one-to-one if it passes the horizontal line test, meaning each horizontal line crosses its graph at most once. The given function is \( f(x) = \sqrt{6 + x} \), where \( x \geq -6 \). Squares with a non-negative radicand are one-to-one because the output rises or stays constant as the input increases. Hence, \( f(x) \) is one-to-one.

Step 2: Find the Equation for the Inverse Function

To find the inverse, start by setting \( y = \sqrt{6 + x} \). Then solve for \( x \) in terms of \( y \): 1. Square both sides: \( y^2 = 6 + x \).2. Isolate \( x \): \( x = y^2 - 6 \). Thus, the inverse function is \( f^{-1}(x) = x^2 - 6 \).

Step 3: Graph the Functions

Graph the function \( f(x) = \sqrt{6 + x} \), which is a transformation of the basic square root function shifted left by 6 units. Also, graph \( f^{-1}(x) = x^2 - 6 \), a transformation of the quadratic \( x^2 \) shifted down by 6 units. The graph of \( f(x) \) is the top half of a sideways parabola starting at (-6, 0). The graph of \( f^{-1}(x) \) is the rightward-opening parabola with a vertex at (0, -6). Both graphs will be reflections over the line \( y = x \).

Step 4: Determine Domain and Range

- For \( f(x) = \sqrt{6 + x} \), the domain is \( x \geq -6 \) and the range is \( y \geq 0 \) because all square root function outputs are non-negative starting from the lowest x value.- For \( f^{-1}(x) = x^2 - 6 \), the domain is \( x \geq 0 \) because it corresponds to the range of \( f(x) \), and the range is \( y \geq -6 \) corresponding to the domain of \( f(x) \).

Key Concepts

One-to-One Functions

One-to-one functions are vital to understanding inverse functions because only they ensure that each output is uniquely paired with an input. This uniqueness is critical when defining inverse functions since each output in the range must correspond precisely to only one input value in the domain.

• A practical way to check if a function is one-to-one is by using the Horizontal Line Test.
• A function is considered one-to-one if no horizontal line intersects the graph more than once.

For the function given, \( f(x) = \sqrt{6 + x} \), it is one-to-one over the domain \( x \geq -6 \). This is because as \( x \) increases, the value of \( f(x) \) continually increases, never repeating any value.

Horizontal Line Test

The Horizontal Line Test is a quick graphical method to determine if a function is one-to-one. If any horizontal line crosses the graph more than once, the function fails the test and is not one-to-one.

• Draw a few horizontal lines through the graph.
• If each line touches the graph at most once, then the function is one-to-one.

For \( f(x) = \sqrt{6 + x} \), each horizontal line will intersect the graph at most once, confirming it passes the test. Thus, the function is one-to-one, allowing us to find an inverse function, which reflects this relationship.

Domain and Range

Understanding the domain and range is crucial when dealing with functions and their inverses. For a function to have an inverse, both the function and the inverse must have their domains and ranges properly defined.

• The domain is the set of all possible inputs for a function.
• The range is the set of all possible outputs.

For \( f(x) = \sqrt{6 + x} \):

• Domain: \( x \geq -6 \). This is due to the square root requiring a non-negative radicand.
• Range: \( y \geq 0 \), as square roots return non-negative numbers.

For its inverse \( f^{-1}(x) = x^2 - 6 \):

• Domain: \( x \geq 0 \), matching with the range of \( f \).
• Range: \( y \geq -6 \), as \( x^2 \) starts from 0 and shifts down by 6.

Graphing Functions

When graphing a function and its inverse, symmetry is a key feature. The original function and its inverse will be mirror images over the line \( y = x \). This is because their roles of input and output are swapped.

• Graph \( f(x) = \sqrt{6 + x} \) as a rightward-extending parabola from the point \(-6, 0 \).
• Graph \( f^{-1}(x) = x^2 - 6 \) as an upward-opening parabola starting at \(0, -6\).

This reflection property helps to understand how changes in input for the original function reflect as changes in output for the inverse. It's a neat visual proof of the function's and its inverse's symmetry and mutual dependency.