What does inverse mean?

What exactly does inverse mean? I've seen it mean the "undoing" function, and also simply meaning "1 over" a function.

Consider the function [tex]f(x)=4x + 6[/tex]
Is the "inverse" [tex]f^{-1}(x)=\frac{x-6}{4}[/tex]
or is it [tex]f^{-1}(x)=\frac{1}{4x+6}[/tex]?

I'm getting kinda confused. It doesn't help that inverse trig functions are annotated with a [tex]^{-1}[/tex], when those are actually the 'undoing' functions.
 

chroot

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In the context of a function, the inverse of a function is another function which "undoes" it.

In general, it is definitely not found by simply dividing one by the function. The easiest way to find the inverse is to solve the function f(x) = x for x, then rename the variables. The first answer you provided is the correct answer.

- Warren
 
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Here is the definition of inverse function from mathworld: Given a function f(x), its inverse [itex]f^{-1}(x)[/itex] is defined by [itex]f(f^{-1}(x)) = f^{-1}(f(x)) \equiv x[/itex] Source: http://mathworld.wolfram.com/InverseFunction.html

So, the inverse of [itex]f(x)=4x + 6[/itex] is what you wrote first: [itex]f^{-1}(x)=\frac{x-6}{4}[/itex] (which chroot explained how to find)

[itex]f^{-1}(x)[/itex] means inverse of f(x), not [tex]\frac{1}{f(x)}[/tex]

So if we compute [itex]f(f^{-1}(x))[/itex]

We get [itex]f(f^{-1}(x)) = f(\frac{x-6}{4}) = 4(\frac{x-6}{4}) + 6 = x - 6 + 6 = x[/itex] which is what we should get, and if we compute [itex]f^{-1}(f(x))[/itex] we should, and will, get x as well.
 
Last edited:

HallsofIvy

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"Inverse", alone, does not ever mean "1 over" a function! That would be the "multiplicative inverse" or reciprocal, just as "negative 1 times" a function would be the "additive inverse" or negative of a function.

The "inverse function" is always the inverse of the composition operation.
 
Hmm, alright, that explains things quite a bit more! So, branching off what's been said, is there a proper way to denote [tex]\frac{1}{f(x)}[/tex] or is that the only way? What about [tex]f(x)[/tex] raised to different powers, such as squared, do you denote this as [tex]f^{2}(x)[/tex]?
 

Alkatran

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KingNothing said:
Hmm, alright, that explains things quite a bit more! So, branching off what's been said, is there a proper way to denote [tex]\frac{1}{f(x)}[/tex] or is that the only way? What about [tex]f(x)[/tex] raised to different powers, such as squared, do you denote this as [tex]f^{2}(x)[/tex]?
Putting a positive exponent after the f usually means "raise the result to the power of. But notations change slightly depending on what you're working with, so it's usually a good idea to write it as:

[tex](f(x))^2[/tex]
 

HallsofIvy

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Yes, it's an unfortunate notation! I think most people would understand [itex]f^{-1}(x)[/itex] to be inverse function, [itex](f(x))^{-1}[/itex] to be reciprocal.

By the way, I teach mathematics in sign language (Gallaudet University in Washington, D.C.). Whenever I introduce the idea of inverse function, I always have to correct students who want to sign it as "reciprocal" (hold out the index and middle fingers horizontally, in a sideways v, the twist your wrist to flip the two fingers over) and insist that the sign as "change" done in reverse. "Change": hold index and thumb of each hand together- the two hands together at your chest- rotate, right hand rotating forward, left backward. "inverse" same hand position, left hand rotates forward, right backward.
 

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