Finding the Inverse of a Function: Examples


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Finding the Inverse of a Function (page 4 of 7)

Sections: Definition / Inverting a graph, Is the inverse a function?, Finding inverses, Proving inverses

Find the inverse of y = x² + 1, x > 0, and determine whether the inverse is a function.

You'll notice that the only difference here is that the domain has been restricted to the positive x-axis this time. Here's the graph:

y = x^2 + 1, x >= 0

Since this passes the Horizontal Line Test, you know that it's inverse will be a function. And since this graph is different from that of the previous function, you know that the inverse must be different. Again, it is very helpful to first find the domains and ranges. The function's domain is
x > 0; the range (from the graph) is y > 1. Then the inverse's domain will be x > 1 and the range will be y > 0. Here's the algebra: Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

The original function:
Solve for "x =":


Since I already figured out the domain and range, I know that I have to choose the positive square root:
Switch the x's and y's; the "y =" is the inverse:

Here's the graph:

function and inverse function

Then the inverse is y = sqrt(x – 1), x > 1, and the inverse is also a function.

Find the inverse of y = ^–2 / _{(x – 5)}, and determine whether the inverse is also a function.

Since the variable is in the denominator, this is a rational function. Here's the algebra:

The original function:
Multiply the denominator up:
Take the x through the parentheses:
Get the x-stuff by itself:
Solve for x:
Switch the x's and y's:

This is just another rational function. The inverse function is y = ^{(5x – 2)} / _x

Find the inverse of f(x) = –sqrt(x – 2), x > 2. Determine whether the inverse is also a function, and find the domain and range of the inverse.

The domain restriction comes from the fact that x is inside a square root. Usually you wouldn't bother writing "x > 2", because you would know that x-values less than 2 would give you negatives inside the square root. But the restriction is useful in this case because, together with the graph, it helps you determine the domain and range on the inverse:

y = -sqrt(x - 2)

The domain is x > 2; the range (from the graph) is y < 0. Then the domain of the inverse will be
x < 0; the range will be y > 2. Here's the algebra:

The original function:
Rename "f(x)" as "y":
Solve for "x =":
Solve for "x =":
Switch x and y:
Rename "y" as "f-inverse". Since I already figured out the domain and range, I know which half of the quadratic I have to choose: