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Restricting the domain

If a function is not one-to-one, it cannot have an inverse. If we restrict the domain of the function so that it becomes one-to-one, thus creating a new function, this new function will have an inverse.

Given a polynomial function, restrict the domain of a function that is not one-to-one and then find the inverse.

  1. Restrict the domain by determining a domain on which the original function is one-to-one.
  2. Replace f ( x ) with y .
  3. Interchange x and y .
  4. Solve for y , and rename the function or pair of function f −1 ( x ) .
  5. Revise the formula for f −1 ( x ) by ensuring that the outputs of the inverse function correspond to the restricted domain of the original function.

Restricting the domain to find the inverse of a polynomial function

Find the inverse function of f :

  1. f ( x ) = ( x 4 ) 2 ,   x 4
  2. f ( x ) = ( x 4 ) 2 ,   x 4

The original function f ( x ) = ( x 4 ) 2 is not one-to-one, but the function is restricted to a domain of x 4 or x 4 on which it is one-to-one. See [link] .

Two graphs of f(x)=(x-4)^2 where the first is when x>=4 and the second is when x<=4.

To find the inverse, start by replacing f ( x ) with the simple variable y .

y = ( x 4 ) 2 Interchange x and  y . x = ( y 4 ) 2 Take the square root . ± x = y 4 Add   4   to both sides . 4 ± x = y

This is not a function as written. We need to examine the restrictions on the domain of the original function to determine the inverse. Since we reversed the roles of x and y for the original f ( x ) , we looked at the domain: the values x could assume. When we reversed the roles of x and y , this gave us the values y could assume. For this function, x 4 , so for the inverse, we should have y 4 , which is what our inverse function gives.

  1. The domain of the original function was restricted to x 4 , so the outputs of the inverse need to be the same, f ( x ) 4 , and we must use the + case:
    f 1 ( x ) = 4 + x
  2. The domain of the original function was restricted to x 4 , so the outputs of the inverse need to be the same, f ( x ) 4 , and we must use the – case:
    f 1 ( x ) = 4 x
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Finding the inverse of a quadratic function when the restriction is not specified

Restrict the domain and then find the inverse of

f ( x ) = ( x 2 ) 2 3.

We can see this is a parabola with vertex at ( 2 , –3 ) that opens upward. Because the graph will be decreasing on one side of the vertex and increasing on the other side, we can restrict this function to a domain on which it will be one-to-one by limiting the domain to x 2.

To find the inverse, we will use the vertex form of the quadratic. We start by replacing f ( x ) with a simple variable, y , then solve for x .

y = ( x 2 ) 2 3    Interchange  x  and  y . x = ( y 2 ) 2 3    Add 3 to both sides . x + 3 = ( y 2 ) 2    Take the square root . ± x + 3 = y 2    Add 2 to both sides . 2 ± x + 3 = y    Rename the function . f 1 ( x ) = 2 ± x + 3

Now we need to determine which case to use. Because we restricted our original function to a domain of x 2 , the outputs of the inverse should be the same, telling us to utilize the + case

f 1 ( x ) = 2 + x + 3

If the quadratic had not been given in vertex form, rewriting it into vertex form would be the first step. This way we may easily observe the coordinates of the vertex to help us restrict the domain.

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Find the inverse of the function f ( x ) = x 2 + 1 , on the domain x 0.

f 1 ( x ) = x 1

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Practice Key Terms 1

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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