Question 380911
f(x) = 9x + 1


let y = f(x)


you get y = 9x + 1


let y = x and x = y


you get x = 9y + 1


solve for y to get y = (x-1)/9


that's your inverse function.


if it is a true inverse function, it will be a reflection of the original function about the line y = x.


to show that, then graph the functions y = x, y = 9x+1, y = (x-1)/9.


that graph is shown below:


{{{graph(600,600,-10,10,-10,10,x,9x+1,(x-1)/9)}}}


you can pretty much eyeball it and see that the graph of y = (x-1)/9 is a reflection of y = 9x+1 about the line y = x.


a more exact interpretation is that f(x,y) in the normal equation should be equal to f(y,x) in the inverse equation.


how you find that out is as follows:


let x = 5 in the normal equation.


then you get y = 9x + 1 which becomes 46


in your normal equation, when x = 5, then y = 46


in your inverse equation, when x = 46, you should get y = 5.


the y in the normal equation becomes the x in the inverse equation.


the x in the normal equation becomes the y in the inverse equation.


when x = 46, y = (x-1)/9 becomes y = (46-1)/9 becomes y = 45/9 becomes 5.


f(x,y) = f(5,46) in your normal equation.


f(y,x) = f(46,5) in your inverse equation.


this proves they are inverse equations.


the inverse equation undoes what the normal equation does.


normal equation takes 5 and makes it 46.


inverse equation takes 46 and makes it 5.