Can you take the inverse of any function?

[ZeroPivot]

i know when you are dealing with limits you can take the inverse to fit the standard limit equations.

how about integrals? can u take the inverse for instance: integral(f(x)dx)

turn it into integral((1/f(x))dx)^-1) get the answer and then reverse it back?

when can u or can't you take the inverse?

 

[HallsofIvy]

First, not every function has an inverse. But if a function has an inverse, then, theoretically, you can find it. Exactly how you would find that inverse and how hard it would be depends on the function.

Do you have some specific reason for asking about the inverse of an integral? In general the inverse of the integral of a function is NOT the integral of the inverse of the function. Your last expression has an unmatched right parenthesis so it is hard to tell exactly what you intend.

 

[ZeroPivot]

lets say the fuction is continuous and has an inverse.

something easy say y=2x+1 and u want the integral integral(2x+1)dx can u switch it around and say (integral(1/(2x+1))dx)^-1, are there any integrals u can do this to or is it a no no?

 

[verty]

lets say the fuction is continuous and has an inverse.

something easy say y=2x+1 and u want the integral integral(2x+1)dx can u switch it around and say (integral(1/(2x+1))dx)^-1, are there any integrals u can do this to or is it a no no?

Integrate that and see what you get. It'll have nothing to do with the inverse of y = 2x+1.

 

[Mandelbroth]

i know when you are dealing with limits you can take the inverse to fit the standard limit equations.

how about integrals? can u take the inverse for instance: integral(f(x)dx)

turn it into integral((1/f(x))dx)^-1) get the answer and then reverse it back?

when can u or can't you take the inverse?

You can take the inverse of any function. (pause for shock value :tongue:)

HOWEVER, it is important to note that the inverse of a function is not necessarily a function. For example, consider the $\sin$ function. Since $\sin(x)=\sin(x+2\pi)$, $\sin^{-1}(\sin(x))$ will not be unique. In fact, $\sin^{-1}(\sin(x))=x+2\pi n$, where $n\in\mathbb{Z}$.