Question 805752
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.
{{{y=x^2}}} is an example. {{{graph(200,200,-3,3,-1,9,x^2)}}}
Each x has one and only one {{{x^2}}},
but the same y value can be the y for more than one x,
for example {{{y=4}}} is the y for {{{x=2}}} and for {{{x=-2}}}.
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)   {{{f(x)=x^2-2x}}} {{{graph(200,200,-2,4,-2,8,x^2-2x)}}}
{{{f(0)=0=f(2)}}} so {{{x=0}}} and {{{x=2}}} have the same {{{y=0}}} ----> There is no inverse function     
(II)   {{{f(x)=1/x}}} can be written as {{{y=1/x}}} {{{graph(200,200,-2,8,-2,8,1/x)}}}
Exchanging the places of the variables we get {{{x=1/y}}} --> {{{y=1/x}}}
So {{{y=1/x}}} or {{{g(x)=1/x)}}} is the inverse function of {{{f(x)=1/x}}}
That function has an inverse, and is its own inverse.          
(III)   {{{f(x)=cos x}}} {{{graph(300,100,-1.5,7.5,-1.5,1.5,cos(x))}}}
We know that cosine is a periodic function with period {{{2pi}}} so the of {{{y=f(x)=cos(x)}}} repeat at {{{2pi}}} intervals so that {{{cos(0)1=cos(2pi)=cos(4pi)}}}{{{"= ...."}}} so we would not know what value of x to assign to y=1.
{{{f(x)=cos x}}} does not have an inverse function 
(IV)   {{{f(x)=sin(x)}}}, {{{-pi/2<=x<=pi/2}}} has a restricted range.
{{{f(-pi/2)=-1}}} and as x increases {{{f(x)=sin x}}} increases, all the way to {{{f(pi/2)=sin (pi/2)=1}}} , without repeating any values.
For {{{pi/2<x<3pi/2}}} the values of {{{sin(x)}}} decrease from 1 all the way to -1, repeating values already found for {{{sin(x)}}} in the {{{(matrix(1,3, -pi/2 ,",", pi/2) )}}} interval.
However, with {{{f(x)}}} defined as {{{f(x)=sin(x)}}} only in the restricted domain {{{-pi/2<=x<=pi/2}}},
each x corresponds to one and only one y, and vice versa.
The function has an inverse.
 
So the answer is {{{highlight((D))}}} II and IV only