Answer
All four pairs of functions \( f \) and \( g \) are inverses of each other. Their graphs are reflections of each other across the line \( y = x \).
Solution
Sure! Let's verify whether each pair of functions \( f \) and \( g \) are inverses of each other using the theorem on inverse functions. We'll also discuss how their graphs relate graphically.
### **Theorem on Inverse Functions**
Two functions \( f \) and \( g \) are inverses of each other if and only if:
1. \( f(g(x)) = x \) for all \( x \) in the domain of \( g \).
2. \( g(f(x)) = x \) for all \( x \) in the domain of \( f \).
When two functions are inverses, their graphs are **reflections** of each other across the line \( y = x \).
Let's apply this to each pair of functions provided.
---
## **1. \( f(x) = \frac{1}{2}x - 3 \) and \( g(x) = 2x + 6 \)**
### **Verification:**
- **Calculate \( f(g(x)) \):**
\[
f(g(x)) = f(2x + 6) = \frac{1}{2}(2x + 6) - 3 = x + 3 - 3 = x
\]
- **Calculate \( g(f(x)) \):**
\[
g\left(\frac{1}{2}x - 3\right) = 2\left(\frac{1}{2}x - 3\right) + 6 = x - 6 + 6 = x
\]
Since both compositions yield \( x \), \( g \) is indeed the inverse of \( f \).
### **Graphical Interpretation:**
- The graph of \( f \) is a straight line with slope \( \frac{1}{2} \) and y-intercept at \( -3 \).
- The graph of \( g \) is a straight line with slope \( 2 \) and y-intercept at \( 6 \).
- These lines are reflections of each other across the line \( y = x \).
---
## **2. \( f(x) = \sqrt[3]{x} - 1 \) and \( g(x) = (x + 1)^3 \)**
### **Verification:**
- **Calculate \( f(g(x)) \):**
\[
f\big(g(x)\big) = f\left((x + 1)^3\right) = \sqrt[3]{(x + 1)^3} - 1 = x + 1 - 1 = x
\]
- **Calculate \( g(f(x)) \):**
\[
g\big(f(x)\big) = g\left(\sqrt[3]{x} - 1\right) = \left(\sqrt[3]{x} - 1 + 1\right)^3 = \left(\sqrt[3]{x}\right)^3 = x
\]
Both compositions yield \( x \), so \( g \) is the inverse of \( f \).
### **Graphical Interpretation:**
- The graph of \( f \) is the cube root function shifted down by 1 unit.
- The graph of \( g \) is the cubic function shifted left by 1 unit.
- These graphs are reflections of each other across the line \( y = x \).
---
## **3. \( f(x) = \frac{x + 2}{3} \) and \( g(x) = 3x - 2 \)**
### **Verification:**
- **Calculate \( f(g(x)) \):**
\[
f(g(x)) = f(3x - 2) = \frac{(3x - 2) + 2}{3} = \frac{3x}{3} = x
\]
- **Calculate \( g(f(x)) \):**
\[
g\left(\frac{x + 2}{3}\right) = 3\left(\frac{x + 2}{3}\right) - 2 = (x + 2) - 2 = x
\]
Since both compositions equal \( x \), \( g \) is the inverse of \( f \).
### **Graphical Interpretation:**
- The graph of \( f \) is a straight line with slope \( \frac{1}{3} \) and y-intercept at \( \frac{2}{3} \).
- The graph of \( g \) is a straight line with slope \( 3 \) and y-intercept at \( -2 \).
- These lines are reflections of each other across the line \( y = x \).
---
## **4. \( f(x) = \sqrt[3]{x + 2} \) and \( g(x) = x^3 - 2 \)**
### **Verification:**
- **Calculate \( f(g(x)) \):**
\[
f\big(g(x)\big) = f(x^3 - 2) = \sqrt[3]{(x^3 - 2) + 2} = \sqrt[3]{x^3} = x
\]
- **Calculate \( g(f(x)) \):**
\[
g\big(f(x)\big) = g\left(\sqrt[3]{x + 2}\right) = \left(\sqrt[3]{x + 2}\right)^3 - 2 = (x + 2) - 2 = x
\]
Both compositions result in \( x \), confirming that \( g \) is the inverse of \( f \).
### **Graphical Interpretation:**
- The graph of \( f \) is the cube root function shifted left by 2 units.
- The graph of \( g \) is the cubic function shifted down by 2 units.
- These graphs are reflections of each other across the line \( y = x \).
---
## **Summary of Graphs**
For each pair of functions \( f \) and \( g \):
- **Verification:** We showed \( f(g(x)) = x \) and \( g(f(x)) = x \), proving that \( g \) is the inverse of \( f \).
- **Graphical Relationship:** The graphs of each pair \( f \) and \( g \) are mirror images across the line \( y = x \), illustrating their inverse relationship.
If you'd like to see the graphs plotted, you can use graphing software or tools like Desmos to visualize these functions and their inverses.
Reviewed and approved by the UpStudy tutoring team
Explain
Simplify this solution