Answer :
Certainly! Let's find the inverse function [tex]\( f^{-1}(x) \)[/tex] for the given function [tex]\( f(x) = \frac{1}{3} x - 1 \)[/tex]. We will then verify that the functions are indeed inverses through composition.
### Step-by-Step Solution:
1. Express [tex]\( y \)[/tex] in Terms of [tex]\( x \)[/tex]:
Start by setting [tex]\( y = f(x) \)[/tex]:
[tex]\[ y = \frac{1}{3} x - 1 \][/tex]
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
To find the inverse function, swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex] in the equation:
[tex]\[ x = \frac{1}{3} y - 1 \][/tex]
3. Solve for [tex]\( y \)[/tex]:
Now, we need to solve this equation for [tex]\( y \)[/tex]:
[tex]\[ x = \frac{1}{3} y - 1 \][/tex]
First, add 1 to both sides to isolate the term involving [tex]\( y \)[/tex]:
[tex]\[ x + 1 = \frac{1}{3} y \][/tex]
Next, multiply both sides by 3 to eliminate the fraction:
[tex]\[ 3(x + 1) = y \][/tex]
Simplify the expression:
[tex]\[ y = 3x + 3 \][/tex]
4. Write the Inverse Function:
Therefore, the inverse function [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[ f^{-1}(x) = 3x + 3 \][/tex]
### Verification by Composition:
To verify that the functions [tex]\( f(x) \)[/tex] and [tex]\( f^{-1}(x) \)[/tex] are inverses, we need to check that [tex]\( f(f^{-1}(x)) = x \)[/tex] and [tex]\( f^{-1}(f(x)) = x \)[/tex].
1. Check [tex]\( f(f^{-1}(x)) = x \)[/tex]:
[tex]\[ f(f^{-1}(x)) = f(3x + 3) \][/tex]
Substitute [tex]\( 3x + 3 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(3x + 3) = \frac{1}{3} (3x + 3) - 1 \][/tex]
Simplify:
[tex]\[ f(3x + 3) = x + 1 - 1 = x \][/tex]
2. Check [tex]\( f^{-1}(f(x)) = x \)[/tex]:
[tex]\[ f^{-1}(f(x)) = f^{-1}\left( \frac{1}{3} x - 1 \right) \][/tex]
Substitute [tex]\( \frac{1}{3} x - 1 \)[/tex] into [tex]\( f^{-1}(x) \)[/tex]:
[tex]\[ f^{-1}\left( \frac{1}{3} x - 1 \right) = 3\left( \frac{1}{3} x - 1 \right) + 3 \][/tex]
Simplify:
[tex]\[ f^{-1}\left( \frac{1}{3} x - 1 \right) = x - 3 + 3 = x \][/tex]
Both compositions return [tex]\( x \)[/tex], confirming that the functions [tex]\( f(x) \)[/tex] and [tex]\( f^{-1}(x) \)[/tex] are indeed inverses.
### Graphing the Function and Its Inverse:
To graph the function [tex]\( f(x) \)[/tex] and its inverse [tex]\( f^{-1}(x) \)[/tex]:
- The function [tex]\( f(x) = \frac{1}{3} x - 1 \)[/tex] is a line with a slope of [tex]\( \frac{1}{3} \)[/tex] and y-intercept of -1.
- The inverse function [tex]\( f^{-1}(x) = 3x + 3 \)[/tex] is a line with a slope of 3 and y-intercept of 3.
Plotting these two lines will show that they are reflections across the line [tex]\( y = x \)[/tex]. This reflection property is a characteristic of inverse functions when graphed.
### Step-by-Step Solution:
1. Express [tex]\( y \)[/tex] in Terms of [tex]\( x \)[/tex]:
Start by setting [tex]\( y = f(x) \)[/tex]:
[tex]\[ y = \frac{1}{3} x - 1 \][/tex]
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
To find the inverse function, swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex] in the equation:
[tex]\[ x = \frac{1}{3} y - 1 \][/tex]
3. Solve for [tex]\( y \)[/tex]:
Now, we need to solve this equation for [tex]\( y \)[/tex]:
[tex]\[ x = \frac{1}{3} y - 1 \][/tex]
First, add 1 to both sides to isolate the term involving [tex]\( y \)[/tex]:
[tex]\[ x + 1 = \frac{1}{3} y \][/tex]
Next, multiply both sides by 3 to eliminate the fraction:
[tex]\[ 3(x + 1) = y \][/tex]
Simplify the expression:
[tex]\[ y = 3x + 3 \][/tex]
4. Write the Inverse Function:
Therefore, the inverse function [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[ f^{-1}(x) = 3x + 3 \][/tex]
### Verification by Composition:
To verify that the functions [tex]\( f(x) \)[/tex] and [tex]\( f^{-1}(x) \)[/tex] are inverses, we need to check that [tex]\( f(f^{-1}(x)) = x \)[/tex] and [tex]\( f^{-1}(f(x)) = x \)[/tex].
1. Check [tex]\( f(f^{-1}(x)) = x \)[/tex]:
[tex]\[ f(f^{-1}(x)) = f(3x + 3) \][/tex]
Substitute [tex]\( 3x + 3 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(3x + 3) = \frac{1}{3} (3x + 3) - 1 \][/tex]
Simplify:
[tex]\[ f(3x + 3) = x + 1 - 1 = x \][/tex]
2. Check [tex]\( f^{-1}(f(x)) = x \)[/tex]:
[tex]\[ f^{-1}(f(x)) = f^{-1}\left( \frac{1}{3} x - 1 \right) \][/tex]
Substitute [tex]\( \frac{1}{3} x - 1 \)[/tex] into [tex]\( f^{-1}(x) \)[/tex]:
[tex]\[ f^{-1}\left( \frac{1}{3} x - 1 \right) = 3\left( \frac{1}{3} x - 1 \right) + 3 \][/tex]
Simplify:
[tex]\[ f^{-1}\left( \frac{1}{3} x - 1 \right) = x - 3 + 3 = x \][/tex]
Both compositions return [tex]\( x \)[/tex], confirming that the functions [tex]\( f(x) \)[/tex] and [tex]\( f^{-1}(x) \)[/tex] are indeed inverses.
### Graphing the Function and Its Inverse:
To graph the function [tex]\( f(x) \)[/tex] and its inverse [tex]\( f^{-1}(x) \)[/tex]:
- The function [tex]\( f(x) = \frac{1}{3} x - 1 \)[/tex] is a line with a slope of [tex]\( \frac{1}{3} \)[/tex] and y-intercept of -1.
- The inverse function [tex]\( f^{-1}(x) = 3x + 3 \)[/tex] is a line with a slope of 3 and y-intercept of 3.
Plotting these two lines will show that they are reflections across the line [tex]\( y = x \)[/tex]. This reflection property is a characteristic of inverse functions when graphed.
