Question Statement Find the domain and range of the inverse function f − 1 ( x ) f ( f − 1 ( x )) = f − 1 ( f ( x )) = x
(i) f ( x ) = 4 x − 3 f ( x ) = x − 5 x f ( x ) = x − 1 x + 2 f ( x ) = x + 2 f ( x ) = x 2 + 6 f ( x ) = x + 4 2 x − 1
Background and Explanation To find the inverse of a function, we swap the roles of x y y Domain of f becomes the Range of f − 1 , and the Range of f becomes the Domain of f − 1 . Finally, we verify the inverse by showing that composing the function with its inverse results in the identity function x
Solution 1. Find Domain and Range:
For f ( x ) = 4 x − 3
Since the domain and range swap for the inverse:
Domain f − 1 = Range f = R Range f − 1 = Domain f = R
2. Find the Inverse Function:
Let y = f ( x ) x = f − 1 ( y ) y y + 3 x f − 1 ( y ) = 4 x − 3 = 4 x = 4 y + 3 = 4 y + 3 y x f − 1 ( x ) = 4 x + 3
3. Verification:
f ( f − 1 ( x )) = 4 ( 4 x + 3 ) − 3 = x + 3 − 3 = x f − 1 ( f ( x )) = 4 ( 4 x − 3 ) + 3 = 4 4 x = x f ( f − 1 ( x )) = f − 1 ( f ( x )) = x
1. Find Domain and Range:
f ( x ) = x − 5 x x = 5
Domain f = R − { 5 } Range f − 1 = R − { 5 }
2. Find the Inverse Function:
Let y = x − 5 x y ( x − 5 ) x y − 5 y x y − x x ( y − 1 ) x f − 1 ( x ) = x = x = 5 y = 5 y = y − 1 5 y = x − 1 5 x x = 1
Domain f − 1 = R − { 1 } Range f = R − { 1 }
3. Verification:
f − 1 ( f ( x )) = x − 5 x − 1 5 ( x − 5 x ) = x − 5 x − ( x − 5 ) x − 5 5 x = 5 5 x = x f ( f − 1 ( x )) = x − 1 5 x − 5 x − 1 5 x = x − 1 5 x − 5 ( x − 1 ) x − 1 5 x = 5 5 x = x
1. Find Domain and Range:
f ( x ) = x − 1 x + 2 x = 1
2. Find the Inverse Function:
Let y = x − 1 x + 2 y ( x − 1 ) x y − y x y − x x ( y − 1 ) x f − 1 ( x ) = x + 2 = x + 2 = y + 2 = y + 2 = y − 1 y + 2 = x − 1 x + 2
Domain f − 1 = R − { 1 } Range f − 1 = R − { 1 }
3. Verification:
f ( f − 1 ( x )) = x − 1 x + 2 − 1 x − 1 x + 2 + 2 = x − 1 x + 2 − x + 1 x − 1 x + 2 + 2 x − 2 = 3 3 x = x f − 1 ( f ( x )) = x
1. Find Domain and Range:
f ( x ) = x + 2 x + 2 ≥ 0
Domain f = [ − 2 , ∞ ) Range f = [ 0 , ∞ )
Therefore:
Domain f − 1 = [ 0 , ∞ ) Range f − 1 = [ − 2 , ∞ )
2. Find the Inverse Function:
Let y = x + 2 y 2 x f − 1 ( x ) = x + 2 = y 2 − 2 = x 2 − 2
3. Verification:
f ( f − 1 ( x )) = ( x 2 − 2 ) + 2 = x 2 = x f − 1 ( f ( x )) = ( x + 2 ) 2 − 2 = x + 2 − 2 = x
1. Find Domain and Range:
f ( x ) = x 2 + 6 x ≥ 0
Domain f = [ 0 , ∞ ) Range f = [ 6 , ∞ )
Therefore:
Domain f − 1 = [ 6 , ∞ ) Range f − 1 = [ 0 , ∞ )
2. Find the Inverse Function:
Let y = x 2 + 6 y − 6 x f − 1 ( x ) = x 2 = y − 6 = x − 6
3. Verification:
f ( f − 1 ( x )) = ( x − 6 ) 2 + 6 = x − 6 + 6 = x f − 1 ( f ( x )) = ( x 2 + 6 ) − 6 = x 2 = x
1. Find Domain and Range:
f ( x ) = x + 4 2 x − 1 x = − 4
2. Find the Inverse Function:
Let y = x + 4 2 x − 1 y ( x + 4 ) x y + 4 y x y − 2 x x ( y − 2 ) x f − 1 ( x ) = 2 x − 1 = 2 x − 1 = − 4 y − 1 = − ( 4 y + 1 ) = y − 2 − ( 4 y + 1 ) = 2 − y 4 y + 1 = 2 − x 4 x + 1
Domain f − 1 = R − { 2 } Range f − 1 = R − { − 4 }
3. Verification:
f ( f − 1 ( x )) = 2 − x 4 x + 1 + 4 2 ( 2 − x 4 x + 1 ) − 1 = 2 − x 4 x + 1 + 4 ( 2 − x ) 2 − x 8 x + 2 − ( 2 − x ) = 9 9 x = x f − 1 ( f ( x )) = 2 − ( x + 4 2 x − 1 ) 4 ( x + 4 2 x − 1 ) + 1 = x + 4 2 x + 8 − 2 x + 1 x + 4 8 x − 4 + x + 4 = 9 9 x = x
Inverse Definition: If y = f ( x ) x = f − 1 ( y ) Domain/Range Relationship: Dom ( f ) = Ran ( f − 1 ) Ran ( f ) = Dom ( f − 1 ) Verification Property: f ( f − 1 ( x )) = x f − 1 ( f ( x )) = x Algebraic Manipulation: Isolating x
Summary of Steps
Identify the Domain of f ( x )
Set y = f ( x ) x y
Swap y x f − 1 ( x )
Determine the Domain of f − 1 (which is the Range of f
Verify the result by calculating the compositions f ( f − 1 ( x )) f − 1 ( f ( x )) x