Question 1092562
<br>One version of the standard way to find the inverse of a function is to switch the x and y in the given function and solve for the new y.<br>
{{{x = 10/5y + 2}}}
{{{x-2 = 2/y}}}
{{{y = 2/(x-2)}}}<br>
In the given function, 0 is excluded from the domain because it would make the denominator 0.  That means 0 is excluded from the range of the inverse function.<br>
You can also see that 0 is not in the range of the inverse function, because if you try to solve
{{{2/(x-2) = 0}}}
you get no solution.<br>
Similarly you can see that 2 is excluded from the range of the given function, because, again, if you try to solve
{{{10/5y + 2 = 2}}}
you will get no solution.  And since 2 is excluded from the range of the given function, it is excluded from the domain of the inverse function.  And we can see that by looking at the inverse function, because x=2 would make the denominator 0.<br>
In summary...<br>
For the given function,
Domain is x not equal to 0
Range is y not equal to 2<br>
For the inverse function,
Domain is x not equal to 2
Range is y not equal to 0