Algebra 2: Determining if inverse of a function is a function, and finding domain and range of inverse?

If you can get the domain and range of the original function, they are just swapped in the inverse (domain becomes range and vice-versa)

OK Raia--- First of all a function that has an inverse must satisfy the horizontal line test....right??

You know how to find the inverse and if you substitute the value for the inverse in for x you should get y = x. So, try a couple equations that you know have inverses and then look at the inverses to see what the domains and ranges are...this will prove it to you.

i don't precisely understand what you advise by utilising finding f. f could in basic terms be the unique equation, f(x) = a million/(x+a million)^2. to locate the inverse, in basic terms turn the x and y coordinates and resolve for y. f(x) = a million/(x+a million)^2 would be re-written as y=a million/(x+a million)^2 Flipping the coordinates supplies us x = a million/(y+a million)^2 Multiply the two aspects by utilising (y+a million)^2 to get x(y+a million)^2 = a million Divide the two aspects by utilising x to get (y+a million)^2 = a million/x sq. root the two aspects to get y + a million = +/- sqrt(a million/x) Subtract one - y = +/- sqrt(a million/x) - a million <----- inverse you'll be sure if the equation is a function or not by utilising graphing it, and passing a vertical line for the time of the graph. If at any factor that the vertical line is going via extra effective than one factor, any factor in any respect, then the equation isn't a function. basically a heads up, something with +/- isn't a function. it is going to fail the vertical line try.