Problem 16 - College Algebra

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. Given \(f(x)=x^{3}-5\) and \(g(x)=\frac{2 x}{1-x}\) a. Find \(f(g(x))\) and \(g(f(x)) .\) b. What does the answer tell us about the relationship between \(f(x)\) and \(g(x) ?\)

Step-by-Step Solution

Answer

The functions \( f(x) \) and \( g(x) \) are not inverses, as \( f(g(x)) \neq x \) and \( g(f(x)) \neq x \).

Step 1: Verify if Functions are One-to-One and Non-Decreasing

We start by checking if each given function can be considered one-to-one and non-decreasing on a specific domain. 1. **Function \( f(x) = x^3 - 5 \):** The function \( x^3 \) is a strictly increasing function for all \( x \), so \( f(x) = x^3 - 5 \) is also strictly increasing for all \( x \). Therefore, \( f(x) \) is one-to-one and non-decreasing over the entire set of real numbers \( \mathbb{R} \).2. **Function \( g(x) = \frac{2x}{1-x} \):** We need to find the domain where it doesn't cause division by zero, which happens when \( x = 1 \). Also, checking the derivative \( g'(x) = \frac{2}{(1-x)^2} \), we find it is positive for all \( x e 1 \), indicating it is increasing. Therefore, \( g(x) \) is one-to-one and non-decreasing on \( (-\infty, 1) \) and \( (1, \infty) \). For simplicity of the inverse, we will choose \( (-\infty, 1) \).

Step 2: Determine the Domain and Find the Inverse of Each Function

Given the established domains, we find the inverses:1. **Inverse of \( f(x) \):** - Switch \( x \) and \( y \) in \( y = x^3 - 5 \), solving for \( x \): \[ x = y^3 - 5 \rightarrow y = \sqrt[3]{x + 5} \] So the inverse is \( f^{-1}(x) = \sqrt[3]{x+5} \).2. **Inverse of \( g(x) \):** - Switch \( x \) and \( y \) in \( y = \frac{2x}{1-x} \), solving for \( x \): \[ x = \frac{2y}{1+y} \] So the inverse is \( g^{-1}(x) = \frac{2x}{1+x} \).3. **Domains:** - \( f(x) \) on \( \mathbb{R} \), inverse \( f^{-1}(x) \) on \( \mathbb{R} \). - \( g(x) \) on \( (-\infty, 1) \), inverse \( g^{-1}(x) \) on \( \mathbb{R} \).

Step 3: Compute f(g(x)) and g(f(x))

Now compute the compositions: 1. **Calculate \( f(g(x)) \):** - Substitute \( g(x) = \frac{2x}{1-x} \) into \( f(x) = x^3 - 5 \): \[ f(g(x)) = \left( \frac{2x}{1-x} \right)^3 - 5 \]2. **Calculate \( g(f(x)) \):** - Substitute \( f(x) = x^3 - 5 \) into \( g(x) = \frac{2x}{1-x} \): \[ g(f(x)) = \frac{2(x^3 - 5)}{1 - (x^3 - 5)} = \frac{2(x^3 - 5)}{6 - x^3} \]

Step 4: Interpret the Results

The compositions \( f(g(x)) eq x \) and \( g(f(x)) eq x \) imply that neither \( f(x) \) nor \( g(x) \) is the inverse of the other over their domains. This tells us that the two functions do not "undo" each other and are not inverses, as the composed results do not return the original input values. However, they remain one-to-one and non-decreasing in their individually defined appropriate domains.

Key Concepts

Domain of a Function

The domain of a function is all the possible input values (usually \(x\) values) for which a function is defined. For every function, it's essential to identify a domain where the function behaves nicely, such as being continuous, one-to-one, or non-decreasing. In the original exercise, it was crucial to find domains where both functions, \(f(x) = x^3 - 5\) and \(g(x) = \frac{2x}{1-x}\), are one-to-one and non-decreasing. This task is important because these characteristics allow the calculation of inverses.

• For \(f(x) = x^3 - 5\), the function \(x^3\) is always strictly increasing, which means \(f(x)\) is one-to-one and non-decreasing across all real numbers \(\mathbb{R}\).
• For \(g(x) = \frac{2x}{1-x}\), it’s vital to exclude the point where the denominator is zero (\(x = 1\)). By examining the derivative, it shows \(g(x)\) is increasing everywhere else, leading to two potential domains: \((-\infty, 1)\) and \((1, \infty)\). However, for inverted function simplicity, \((-\infty, 1)\) was chosen.

Identifying domains ensures the functions are primed for analysis, such as computing inverses.

Composition of Functions

Composition of functions involves plugging one function into another. We write this as \(f(g(x))\) or \(g(f(x))\). To understand whether two functions act as inverses, computing these compositions is a significant step. If the result is simply \(x\), they undo each other perfectly within their composed domains.

• In this case, \(f(g(x))\) equals \(\left(\frac{2x}{1-x}\right)^3 - 5\), which does not equal \(x\).
• Similarly, \(g(f(x))\) results in \(\frac{2(x^3 - 5)}{6 - x^3}\), which also is not \(x\).

These compositions demonstrate that \(f(x)\) and \(g(x)\) do not act as inverses for each other. Instead, they offer a revealing insight, confirming that while they maintain their own one-to-one character within their individual domains, they do not 'undo' each other's effect.

One-to-One Function

A one-to-one function is a function where each input has a unique output, and each output is from one unique input. This property is crucial when it comes to finding the inverse of a function, as only one-to-one functions have inverses that are also functions.

• For the function \(f(x) = x^3 - 5\), since \(x^3\) is strictly increasing, \(f(x)\) is one-to-one over the entire set of real numbers.
• Similarly, the function \(g(x) = \frac{2x}{1-x}\) is verified as one-to-one on the domain \((-\infty, 1)\).

Identifying these characteristics allows the creation of inverses, ensuring that each point has a distinct counterpart in the inverse mapping, making operations reversible only through mutual mapping.

Monotonic Function

Monotonic functions are functions that either never increase or never decrease. Strictly speaking, they are either entirely non-decreasing or non-increasing, over a specified domain. This property is vital for determining if a function is one-to-one.

• The function \(f(x) = x^3 - 5\) is an example of a monotonic function because \(x^3\) is naturally strictly increasing for any real number.
• For \(g(x) = \frac{2x}{1-x}\), by examining the derivative \(g'(x) = \frac{2}{(1-x)^2}\), we find it always positive within its domain, specifically \((-\infty, 1)\). Consequently, the function strictly increases.

Recognizing monotonicity helps in defining invertible functions since these features indicate there are no repeated outputs for different inputs, streamlining the creation of functional inverses.