# Inverse Functions: Justifying the Claim

• rsnd
In summary, inverse functions are functions that are their own inverse. This can be proven algebraically by computing f(f(x)) and graphically by observing the symmetry of the graphs of the function and its inverse. The function y=sin x has an inverse function of y=arcsin x or sin^{-1} x.
rsnd
Inverse functions?

I was just wondering...Isnot y=x the only function that is the inverse of itself? How do you go on about saying it in a formal way? I mean as response to your assignment question. where you are not just told to not only give fatcts but "Justify your claim with appropriate proofs!"?

Thanks
K. Cv.

There's a nice graphic-related proof.A function that is invertible on a domain is bijective,which means both injective & surjective.In terms of graphs,the graph of the function and its inverse (if it exists) are symmetrical wrt the graph of f(x)=x.If the two functions coincide,then they coincide with the identity function.

Daniel.

Do you mean to say, y=x is the only function where $$f(x)=f^{-1}(x)$$?

Proof by counterexample:

$$y=\frac{1}{x}$$

Any function which is symmetrical from the line y=x is the inverse of itself, in that sense there will be an infinite number of them.

Although perhaps very few simple functions, another easy one is $$y^2 + x^2 = r^2$$

The identity function is invertible on all R...

Daniel.

how about y = -x, y = -1/x,...

If your question was simply "prove of disprove: y = x is the only function that is its own inverse", you can just give a counterexample and be done. Is this how the question you were given was worded?

precalculus question

Hello,

I have a similar question; I'm currently taking precalculus, and I have a question here that I'm stuck on:

"Let f(x)=(ax+b)/(cx-a), (cx-a)?0. Show that f(x) is it's own inverse."

I understand, graphically, that f(x) equals its inverse in this case, because they are reflections of each other over y=x. I've graphically solved this using technology and by-hand sketches. And I also understand that the restiction on the denominator come about because the equation would then be undefined (if it equalled 0). But I cannot figure out why this is, algebraically.

Can anyone help me, or give me a hint?

Thanks very much,
sam.grotz

sam.grotz said:
Hello,

I have a similar question; I'm currently taking precalculus, and I have a question here that I'm stuck on:

"Let f(x)=(ax+b)/(cx-a), (cx-a)?0. Show that f(x) is it's own inverse."

I understand, graphically, that f(x) equals its inverse in this case, because they are reflections of each other over y=x. I've graphically solved this using technology and by-hand sketches. And I also understand that the restiction on the denominator come about because the equation would then be undefined (if it equalled 0). But I cannot figure out why this is, algebraically.

Can anyone help me, or give me a hint?

Thanks very much,
sam.grotz

Just compute f(f(x)).

sam.grotz said:
Hello,

I have a similar question; I'm currently taking precalculus, and I have a question here that I'm stuck on:

"Let f(x)=(ax+b)/(cx-a), (cx-a)?0. Show that f(x) is it's own inverse."

I understand, graphically, that f(x) equals its inverse in this case, because they are reflections of each other over y=x. I've graphically solved this using technology and by-hand sketches. And I also understand that the restiction on the denominator come about because the equation would then be undefined (if it equalled 0). But I cannot figure out why this is, algebraically.

Can anyone help me, or give me a hint?

Thanks very much,
sam.grotz

Sam, just start with $$y = \frac{ax+b}{cx-a}$$ and then rearrage it to get "x" by itself on the LHS.

Whats happening. Are you just getting stuck on the algebra to do this rearrangement?

Zurtex said:
Any function which is symmetrical from the line y=x is the inverse of itself, in that sense there will be an infinite number of them.

Although perhaps very few simple functions, another easy one is $$y^2 + x^2 = r^2$$
$$y^2 + x^2 = r^2$$ is not a function, because for any value of x < r, you can get two different values of y.

A silly conjecture... An infinite number of counterexamples.

Thanks!

Thanks to ZioX and uart for giving me the helpful direction. I think what I must have done before was just think too far ahead, and presume that it couldn't be simplified in that way. Thanks again! It turns out it actually ends up simplifying to just:

y=x

peace,
sam.grotz

what is the inverse function of sin x. I don't know if it is written correctly

arcsin, or sin^{-1}

and if you think that it is avoiding the issue, then you need to rethink your view of what a function is.

## 1. What is an inverse function?

An inverse function is a mathematical operation that "undoes" another function. It essentially switches the input and output values of the original function. For example, if a function f(x) takes in a number and multiplies it by 2, the inverse function would take that result and divide it by 2 to get back to the original input value.

## 2. How do you prove that two functions are inverses of each other?

To prove that two functions are inverses, you need to show that when you compose them together (i.e. plug one function's output into the other function), you get back to the original input value. In other words, if f(g(x)) = x and g(f(x)) = x, then these two functions are inverses of each other.

## 3. Can all functions have an inverse?

No, not all functions have an inverse. For a function to have an inverse, it needs to pass the "horizontal line test". This means that if you were to draw a horizontal line through the graph of the function, it should only intersect the graph once. If it intersects more than once, then the function does not have a unique inverse.

## 4. How can inverse functions be useful in real-life applications?

Inverse functions can be used to solve many real-life problems, such as finding the original amount of a discounted price, calculating compound interest, or converting between different units of measurement. They can also be used in engineering, physics, and other scientific fields to model and understand complex systems.

## 5. Is the inverse of a function always a function?

Yes, the inverse of a function is always a function. This is because for a function to have an inverse, it needs to pass the "vertical line test", which means that no vertical line can intersect the graph of the function more than once. This ensures that the inverse function is also a one-to-one mapping, just like the original function.

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