can someone please help me with functions?

Question

let f be a real-valued function with domain D and range R..what are the necessary and sufficient conditions for an inverse function for f to exist? also how do i prove that f can have at most one inverse function?


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also, to define the inverse sine function arcsin x the domain of sin x was restricted to [-π/2, π/2]. why won't this domain work for cos x if we wish to define the inverse function?
if i instead let g(x)=cos x, 0<=x<=π, how do i denote arccos x = g^-1(x)? and what is the domain and range of y=arccos x??


<= means less than or equal to.

thanks for any help with this

Answer 1

If g is the inverse of f, then this condition must be met:

f(g(x))=g(f(x))=x

g composed with f must equal the identity function; there is only one solution per function which does this (so you just need to prove that).


As for the trig stuff, I'm afraid I'm not completely clear what you mean...

Answer 2

Call the original function f and its inverse r. By the definition
1. f: (x,y) within domain D.
and
2. r: (y,x) within its domain R.

3. For f you must not have x1=x2 and y1<>y2.
4. However, you might have x1<>x2 and y1=y2.

If you apply 4 to the inverse r then it violates 3.

5. Therefore for f to have r it is necessary that for x1=x2, y1=y2.

Is it sufficient? That means given f with x1=x2 you have y1=y2, can you find r with y1=y2 for which x1<>x2.
Since from previous argument x1=x2 always leads to y1=y2 this is impossible. That proves it is sufficient.
Finally, is r unique? For f: (x1,y1) the r: (y1,x1). Assume there exist r2: (y2,x2) where x2<>x1 for y1=y2. This also violates 5. which proves r is unique inverse of f.

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Applying it to sin and cos you see the condition 5. is fulfilled for sin within domain -pi/2 <= x <= pi/2, and for cosine within domain 0 <= x <= pi. Therefore, domain for arccos(x) is -1 <= x <= +1 and range pi => y >= -pi.