Answer :
Sure, let's find the inverse functions for each given function and verify them by composition.
### Function [tex]\( f(x) = -2x + 3 \)[/tex]
1. Find the inverse function:
To find the inverse function, we start by replacing [tex]\( f(x) \)[/tex] with [tex]\( y \)[/tex]:
[tex]\[ y = -2x + 3 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ y - 3 = -2x \implies x = \frac{3 - y}{2} \][/tex]
Replace [tex]\( y \)[/tex] with [tex]\( x \)[/tex] to get the inverse function:
[tex]\[ f^{-1}(x) = \frac{3 - x}{2} \][/tex]
2. Verify by composition:
To verify that [tex]\( f^{-1}(x) \)[/tex] is indeed the inverse of [tex]\( f(x) \)[/tex], we need to check that:
[tex]\[ f(f^{-1}(x)) = x \quad \text{and} \quad f^{-1}(f(x)) = x \][/tex]
- Verify [tex]\( f(f^{-1}(x)) \)[/tex]:
[tex]\[ f\left( f^{-1}(x) \right) = f\left( \frac{3 - x}{2} \right) = -2 \left( \frac{3 - x}{2} \right) + 3 = - (3 - x) + 3 = x \][/tex]
- Verify [tex]\( f^{-1}(f(x)) \)[/tex]:
[tex]\[ f^{-1}(f(x)) = f^{-1}(-2x + 3) = \frac{3 - (-2x + 3)}{2} = \frac{3 + 2x - 3}{2} = x \][/tex]
3. Graph:
To graph both [tex]\( f(x) \)[/tex] and [tex]\( f^{-1}(x) \)[/tex], plot the lines:
[tex]\[ f(x) = -2x + 3 \quad \text{and} \quad f^{-1}(x) = \frac{3 - x}{2} \][/tex]
The line [tex]\( y = x \)[/tex] will also be plotted as a reference as it represents the reflection line for the inverse functions.
### Function [tex]\( f(x) = 2x - 3 \)[/tex]
1. Find the inverse function:
Replace [tex]\( f(x) \)[/tex] with [tex]\( y \)[/tex]:
[tex]\[ y = 2x - 3 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ y + 3 = 2x \implies x = \frac{y + 3}{2} \][/tex]
Replace [tex]\( y \)[/tex] with [tex]\( x \)[/tex] to get the inverse function:
[tex]\[ f^{-1}(x) = \frac{x + 3}{2} \][/tex]
2. Verify by composition:
To verify that [tex]\( f^{-1}(x) \)[/tex] is indeed the inverse of [tex]\( f(x) \)[/tex], we need to check that:
[tex]\[ f(f^{-1}(x)) = x \quad \text{and} \quad f^{-1}(f(x)) = x \][/tex]
- Verify [tex]\( f(f^{-1}(x)) \)[/tex]:
[tex]\[ f\left( f^{-1}(x) \right) = f\left( \frac{x + 3}{2} \right) = 2 \left( \frac{x + 3}{2} \right) - 3 = x + 3 - 3 = x \][/tex]
- Verify [tex]\( f^{-1}(f(x)) \)[/tex]:
[tex]\[ f^{-1}(f(x)) = f^{-1}(2x - 3) = \frac{(2x - 3) + 3}{2} = \frac{2x}{2} = x \][/tex]
3. Graph:
To graph both [tex]\( f(x) \)[/tex] and [tex]\( f^{-1}(x) \)[/tex], plot the lines:
[tex]\[ f(x) = 2x - 3 \quad \text{and} \quad f^{-1}(x) = \frac{x + 3}{2} \][/tex]
Again, the line [tex]\( y = x \)[/tex] will be plotted as a reference.
These steps provide clear guidance on finding the inverse functions for [tex]\( f(x) = -2x + 3 \)[/tex] and [tex]\( f(x) = 2x - 3 \)[/tex], verifying them through composition, and graphing the functions along with their inverses.
### Function [tex]\( f(x) = -2x + 3 \)[/tex]
1. Find the inverse function:
To find the inverse function, we start by replacing [tex]\( f(x) \)[/tex] with [tex]\( y \)[/tex]:
[tex]\[ y = -2x + 3 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ y - 3 = -2x \implies x = \frac{3 - y}{2} \][/tex]
Replace [tex]\( y \)[/tex] with [tex]\( x \)[/tex] to get the inverse function:
[tex]\[ f^{-1}(x) = \frac{3 - x}{2} \][/tex]
2. Verify by composition:
To verify that [tex]\( f^{-1}(x) \)[/tex] is indeed the inverse of [tex]\( f(x) \)[/tex], we need to check that:
[tex]\[ f(f^{-1}(x)) = x \quad \text{and} \quad f^{-1}(f(x)) = x \][/tex]
- Verify [tex]\( f(f^{-1}(x)) \)[/tex]:
[tex]\[ f\left( f^{-1}(x) \right) = f\left( \frac{3 - x}{2} \right) = -2 \left( \frac{3 - x}{2} \right) + 3 = - (3 - x) + 3 = x \][/tex]
- Verify [tex]\( f^{-1}(f(x)) \)[/tex]:
[tex]\[ f^{-1}(f(x)) = f^{-1}(-2x + 3) = \frac{3 - (-2x + 3)}{2} = \frac{3 + 2x - 3}{2} = x \][/tex]
3. Graph:
To graph both [tex]\( f(x) \)[/tex] and [tex]\( f^{-1}(x) \)[/tex], plot the lines:
[tex]\[ f(x) = -2x + 3 \quad \text{and} \quad f^{-1}(x) = \frac{3 - x}{2} \][/tex]
The line [tex]\( y = x \)[/tex] will also be plotted as a reference as it represents the reflection line for the inverse functions.
### Function [tex]\( f(x) = 2x - 3 \)[/tex]
1. Find the inverse function:
Replace [tex]\( f(x) \)[/tex] with [tex]\( y \)[/tex]:
[tex]\[ y = 2x - 3 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ y + 3 = 2x \implies x = \frac{y + 3}{2} \][/tex]
Replace [tex]\( y \)[/tex] with [tex]\( x \)[/tex] to get the inverse function:
[tex]\[ f^{-1}(x) = \frac{x + 3}{2} \][/tex]
2. Verify by composition:
To verify that [tex]\( f^{-1}(x) \)[/tex] is indeed the inverse of [tex]\( f(x) \)[/tex], we need to check that:
[tex]\[ f(f^{-1}(x)) = x \quad \text{and} \quad f^{-1}(f(x)) = x \][/tex]
- Verify [tex]\( f(f^{-1}(x)) \)[/tex]:
[tex]\[ f\left( f^{-1}(x) \right) = f\left( \frac{x + 3}{2} \right) = 2 \left( \frac{x + 3}{2} \right) - 3 = x + 3 - 3 = x \][/tex]
- Verify [tex]\( f^{-1}(f(x)) \)[/tex]:
[tex]\[ f^{-1}(f(x)) = f^{-1}(2x - 3) = \frac{(2x - 3) + 3}{2} = \frac{2x}{2} = x \][/tex]
3. Graph:
To graph both [tex]\( f(x) \)[/tex] and [tex]\( f^{-1}(x) \)[/tex], plot the lines:
[tex]\[ f(x) = 2x - 3 \quad \text{and} \quad f^{-1}(x) = \frac{x + 3}{2} \][/tex]
Again, the line [tex]\( y = x \)[/tex] will be plotted as a reference.
These steps provide clear guidance on finding the inverse functions for [tex]\( f(x) = -2x + 3 \)[/tex] and [tex]\( f(x) = 2x - 3 \)[/tex], verifying them through composition, and graphing the functions along with their inverses.
