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Operations with Functions
  
 
Inverse Functions
Inverse Functions
Two functions f and g are inverse functions if fog(x) = x and gof (x) = x for all values of x in the domain of f and g.
For instance, f (x) = 2x and g(x) = x are inverse functions because fog(x) = f (g(x)) = f (x) = 2(x) = x and gof (x) = g(f (x)) = g(2x) = (2x) = x. Similarly, f (x) = x + 1 and g(x) = x - 1 are inverse funcions because fog(x) = f (g(x)) = f (x - 1) = (x - 1) + 1 = x and gof (x) = g(f (x)) = g(x + 1) = (x + 1) - 1 = x. h(x) = 3x - 1 and j(x) = are inverse functions because hoj(x) = h(j(x)) = h() = 3() - 1 = x + 1 - 1 = x and joh(x) = j(h(x)) = j(3x - 1) = = = x.
The inverse of a function f (x) is denoted f-1(x).
Finding the Inverse of a Function by Reversing Operations
The trick to finding the inverse of a function f (x) is to "undo" all the operations on x in reverse order.
The function f (x) = 2x - 4 has two steps:
  1. Multiply by 2.
  2. Subtract 4.
Thus, f-1(x) must have two steps:
  1. Add 4.
  2. Divide by 2.
Consequently, f-1(x) = .
We can verify that this is the inverse of f (x):
f-1(f (x)) = f-1(2x - 4) = = = x.

f (f-1(x)) = f () = 2() - 4 = (x + 4) - 4 = x.

Example 1: Find the inverse of f (x) = 3(x - 5).

Original function:
  1. Subtract 5.
  2. Multiply by 3.
New function:
  1. Divide by 3.
  2. Add 5.
Thus, f-1(x) = + 5.

Check:
f-1(f (x)) = f-1(3(x - 5)) = + 5 = (x - 5) + 5 = x.
f (f-1(x)) = f ( +5) = 3(( +5) - 5) = 3() = x.

Example 2: Find the inverse of f (x) = , x≥2 (we must restrict the domain because f (x) is undefined for x < 2).

Original function:
  1. Subract 2.
  2. Take the square root.
New function:
  1. Square.
  2. Add 2.
Thus, f-1(x) = x2 + 2.

Check:
f-1(f (x)) = f-1() = ()2 + 2 = (x - 2) + 2 = x.
f (f-1(x)) = f (x2 +2) = = = x.
When we take the inverse of a function, the domain and range switch. In example 2, the domain of f is x≥2 and the range of f is f (x)≥ 0. Thus, the domain of f-1 is x≥ 0 and the range of f-1 is f-1(x)≥2.
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