SOLUTION: Determine which function(s) below have an inverse f^(-1). Can you please explain to me why? (I) f(x)=x^2-2x (II) f(x)=1/x (III) f(x)=cos x (IV)

Question 805752: Determine which function(s) below have an inverse f^(-1). Can you please explain to me why?
(I) f(x)=x^2-2x (II) f(x)=1/x (III) f(x)=cos x (IV) f(x)=sin x,-π/2≤x≤π/2
(A) I & II
(B) I, II, & III
(C) II, III, & IV
(D) II & IV
(E) none of the above
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
A relation between x and y establishes y as a function of x if for each x we find no more that one y paired to that x.
It could be that y is a function of x that happens to assign the same value of y to more than one x.
is an example.
Each x has one and only one ,
but the same y value can be the y for more than one x,
for example is the y for and for .
In that case, we cannot reverse the function and sat that x is established as a function of y, because for y=4 we find two corresponding values of x.
If there is just one x (or less) for each y, then we can solve for x and find the reverse function.

(I)

so and have the same ----> There is no inverse function
(II) can be written as

Exchanging the places of the variables we get -->
So or is the inverse function of
That function has an inverse, and is its own inverse.
(III)

We know that cosine is a periodic function with period so the of repeat at intervals so that so we would not know what value of x to assign to y=1.
does not have an inverse function
(IV) , has a restricted range.
and as x increases increases, all the way to , without repeating any values.
For the values of decrease from 1 all the way to -1, repeating values already found for in the interval.
However, with defined as only in the restricted domain ,
each x corresponds to one and only one y, and vice versa.
The function has an inverse.

So the answer is II and IV only
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