Problem 15 - College Algebra with Corequisite Support

Question

For the following exercises, find a domain on which each function \(f\) is one- to-one and non-decreasing. Write the domain in interval notation. Then find the inverse of \(f\) restricted to that domain. \(f(x)=x^{2}-5\)

Step-by-Step Solution

Answer

Domain: [0, ∞). Inverse: \( f^{-1}(y) = \sqrt{y + 5} \).

Step 1: Understand the Function

The given function is \( f(x) = x^2 - 5 \). This is a quadratic function, and its graph is a parabola opening upwards. This function is not one-to-one on its entire domain \( \mathbb{R} \) because it is symmetric about the y-axis.

Step 2: Determine a Suitable Domain

To make the quadratic function one-to-one, we need to restrict its domain to where it is either non-decreasing or non-increasing. Because the vertex of the parabola is at \( x=0 \), the function is non-decreasing on the interval \([0, \infty)\). On this interval, the function is increasing, ensuring that it is one-to-one.

Step 3: Express the Domain in Interval Notation

For the function to be one-to-one, we restrict the domain to \([0, \infty)\) where the function is non-decreasing.

Step 4: Find the Inverse Function

To find the inverse of \( f(x) = x^2 - 5 \), solve the equation \( y = x^2 - 5 \) for \( x \). First, add 5 to both sides: \( y + 5 = x^2 \). Then, take the square root of both sides: \( x = \sqrt{y + 5} \). Since we are restricting \( x \geq 0 \), we take the positive root.

Step 5: Write the Inverse Function

The inverse function, using the restricted domain \([0, \infty)\), is \( f^{-1}(y) = \sqrt{y + 5} \). This is valid because the inverse will map values from \([-5, \infty)\) back to \([0, \infty)\) as per the original restricted domain.

Key Concepts

Domain Restriction

Domain restriction is essential when dealing with functions that are not one-to-one across their entire domain, such as quadratic functions. A function must be one-to-one for its inverse to exist. For a given function like \[ f(x) = x^2 - 5 \]we begin by observing that it is not one-to-one over its whole domain of \( \mathbb{R} \).This is evident because the function can produce the same output for two different inputs. In this case, the function's graph is a parabola, symmetrical about the y-axis.
To ensure this function is one-to-one, and to find its inverse, we must perform a domain restriction. For a quadratic function like this, we usually restrict the domain to one side of the vertex, as the graph on either side is monotonous (either non-decreasing or non-increasing). For \( f(x) = x^2 - 5 \), the function is non-decreasing on \([0, \infty)\). Thus, we restrict the domain to this interval, allowing the function to pass the horizontal line test and become one-to-one.

Quadratic Functions

Quadratic functions are polynomial functions of degree two, generally represented as \( f(x) = ax^2 + bx + c \). They have versatile properties but often present challenges when analyzing for inverse functions due to their parabolic shape. A key feature of quadratic functions is the vertex, the point at which the function reaches a maximum or minimum value, depending on whether the parabola opens upwards or downwards.
For example, in the quadratic function \( f(x) = x^2 - 5 \), the parabola opens upwards with a vertex at \( (0, -5) \). This vertex divides the parabola into two sections:

• The section where the graph is decreasing (for \( x < 0 \))
• The section where the graph is increasing (for \( x > 0 \))

Understanding these properties helps determine a suitable domain for inversion. By restricting the domain to the non-decreasing portion, we ensure the function is one-to-one and invertible.

One-to-One Functions

A one-to-one function is a function where each output is paired with exactly one input. This characteristic is crucial when determining the inverse of a function. If a function is one-to-one, it passes the horizontal line test, meaning no horizontal line intersects the graph more than once.
The function \( f(x) = x^2 - 5 \) over the restricted domain \([0, \infty)\) qualifies as one-to-one. This is because, for each value of \( x \)in the given interval, there’s a unique \( f(x) \). With this restriction, the function transforms into an invertible form. Finding the inverse entails solving for \( x \) in terms of \( y \). This gives us the inverse function \( f^{-1}(y) = \sqrt{y + 5} \), valid for \( y \geq -5 \). Each \( y \) in this range corresponds to a unique \( x \), maintaining the one-to-one nature needed for inversion.