Finding the Inverse of a Function with a Trigonometric Term

[Joe_K]

Homework Statement

f(x)= x+cosx

find the inverse, f^-1(x)

The Attempt at a Solution

To start, I tried to solve the original equation for x. But this is where I am having trouble. How do you get x by itself when it is trapped within the cos function? I used to know how to do this but I seem to have forgotten. Once I solve the equation for x, I should be able to switch the "x" and "y" terms and be left with the inverse of the original function. Maybe someone can help me remember how to get the 'x' by itself. Thank you.

 

[LCKurtz]

Your memory of how to do it is fine, but you can't solve this one for x algebraically.

 

[Joe_K]

What do I need to do, in order to do the problem correctly? The question in my book is asking to find f^-1(1). Should I just plug in '1' for y in the equation and solve? I believe that would only leave zero as a possible answer?

 

[LCKurtz]

What do I need to do, in order to do the problem correctly? The question in my book is asking to find f^-1(1). Should I just plug in '1' for y in the equation and solve? I believe that would only leave zero as a possible answer?

Undoubtedly the reason they asked that simpler question is because you can't do the general one. So, yes. Obviously x = 0 is a value solving 1 = x + cos(x). Do you see how to show it is the only solution or, for that matter, that the inverse function exists?

 

[Joe_K]

Yes, thank you for your help!

 

[Dick]

Homework Statement

f(x)= x+cosx

find the inverse, f^-1(x)

The Attempt at a Solution

To start, I tried to solve the original equation for x. But this is where I am having trouble. How do you get x by itself when it is trapped within the cos function? I used to know how to do this but I seem to have forgotten. Once I solve the equation for x, I should be able to switch the "x" and "y" terms and be left with the inverse of the original function. Maybe someone can help me remember how to get the 'x' by itself. Thank you.

I don't think you are going to find a nice expression for f^(-1)(x). You can probably answer some questions about f^(-1) without having an expression for it. Is that the whole question?