f(x) where x belongs to all real numbers inverse: f-1(x), where x belongs to all real numbers True or False: The inverse of f(x+3) is f-1(x+3) My ideas: I think that it is false given that when you usually find the inverse of a function, you switch the x and y variables and solve for y again meaning that the inverse couldn't stay the same. I figured since the domain and range of f(x) belong to all real numbers, possibly f(x) = x and then inputting f(x+3) = x+ 3 then y = x+3 then y = x - 3 but im not really sure if thats right :s
You are given that ##f## has an inverse ##f^{-1}##. What happens when you solve the equation ##y=f(x+3)## for ##x##?
Good Day michellemich! If you are not sure of your answer, try some composition: let your original function be f(x)and your questionable inverse function be g(x) Evaluate (f of g) and (g of f). If they undo each other, they are inverses.
If you want to know if this is true for all invertible functions, it is simple enough to find a counterexample. If, say, f(x)= 2x+ 3, then [itex]f(x)= 3x- 2[/itex], then [itex]f^{-1}(x)= (x+ 2)/3[/itex]. f(x+3)= 3(x+ 3)- 2= 3x+ 7. The inverse of that function is [itex](x- 7)/3[/itex]. Is that equal to [itex]f^{-1}(x+ 3)= (x+3+ 2)/3= (x+ 5)/3[/itex]?