Bob A. answered • 11/21/14
20 Years Making Science and Maths Understandable and Interesting!
To find the inverse f-1(x) of f(x)
First, if you can graph it you can check if it even has an inverse.
To have an inverse a function must pass both
the Vertical Line Test (VLT)
and the Horizontal Line Test (HLT).
For a function f(x). f(x) = -12 x^3
1st you replace the f(x) with y y = -12 x^3
2nd you swap the x and y x = -12 y^3
3rd you solve for the y x/-12 = y^3
(x/-12)1/3 = y
4th you put back
the function notation f-1(x) = (x/-12)1/3
5) The domain of f(x) is the range of f-1(x)
The range of f(x) is the domain of f-1(x)
6) Now those might not really be inverse functions of each other.
To test that they are really inverses of each other you need to
find both composite functions f(f-1(x)) and f-1(f(x)).
Both of these Must = x or else they are not inverses of each other.
7) f(f-1(x)) = -12( (x/-12)1/3 )3
= -12 (x/-12) = x
8) f-1(f(x)) = ( -12 x3) /-12)1/3
=( x3 )1/3 = x
9) Okay, both composites equal x so
they are inverse functions of each other.
Bob A.
11/26/14
Ben B.
At the algebra level, this may be exactly how you will be expected to show all of the steps, or it may be a little different. You should check with your instructor, text, or notes as applicable to understand exactly what steps are necessary to get full credit, specifically with regard to proving that the function has an inverse, on what interval, and what notation, methods and terminology is acceptable. The solution below in my response is equivalent to the 3rd step above, without any swapping of parameters (x to y and y to x), which I think is confusing as it also swaps the meaning of x and y. This 3rd step is the key step for actually determining the candidate inverse function - solving the equation backwards. The choice of parameters is generally arbitrary for functions, depending on the context. What is important is to keep the domains and ranges straight in the event that the inverse exists only for a limited domain (input values). In order to have an inverse, there must be only one y value for every x from the y= f(x) function, and only one x value for every y (using the definition of x and y prior to swapping). In practical applications, it is more typical to identify one parameter for the domain (input) of the original function and a different parameter for the output (range) of the original function, as these may have very different physical meaning, including units and you don't want to confuse the two. It is also important to understand what is happening when you exchange the x and y labels as shown above – this results in a re-definition for the parameter 'x' - as it is being define as the input to f(x), but then re-defined to be the input to the inverse function - a different function pulling from (potentially) a different set of allowed values. This will become more evident when you start looking at functions such as y=cosΘ, where you don’t really want to swap the y with the theta, and the inverse function exists only for limited values of theta (e. g. 0 to pi). All that being said, what Robert has posted is probably better and certainly a more thorough solution using algebra II notation. Thanks, and please let us know if you need any additional clarification. - Ben
11/24/14