# SOLUTION: two function f and g are defined by : f(x)=2x+1 g(x)=x^2 verify that (fog)^-1 = g^-1 o f^-1

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 Question 629914: two function f and g are defined by : f(x)=2x+1 g(x)=x^2 verify that (fog)^-1 = g^-1 o f^-1Answer by jsmallt9(3296)   (Show Source): You can put this solution on YOUR website!Inverses of of functions are found by swapping the x's and y's. To perform this swap, I like to temporarily replace the function notation with a "y": f(x) = 2x+1 y = 2x + 1 Swap the x's and y's: x = 2y + 1 <=== This is the inverse, but not in a desired form Solve for y: x - 1 = 2y Divide by 2 (or multiply by 1/2): Doing the same for g(x): + Note: Because of the +, the inverse of g is not a function (i.e. some, in fact all, x's have more than one y value). And since this inverse is not a function, I do not believe it is proper to use function notation, g^-1 on it. For the compositions it helps if we have a good understanding of what the function's definition is telling you. For example f(x) = 2x+1 The left side tells us that the input to the function, between the parentheses after the function name, is being called "x". "x" is just a placeholder. It is just being used as a name for the input to the function. We could use any letter here to name the input. f(q) = 2q+1 is the exact same function as f(x)!! The right side tells us what function f does with its input. It multiplies the input, x, by 2 and then adds 1. Since "x" is just a place holder, function f will take any input, multiply it by 2 and then add 1: f(7) = 2(7) + 1 f(1002.4) = 2(1002.4) + 1 f(12x-3) = 2(12x-3) + 1 etc. (fog)(x) is just another way to write f(g(x)). In f's parentheses we have "g(x)". So "g(x)" is the input to f. And what does f do to its input? Answer: it multiplies by 2 and then adds 1!: f(g(x)) = 2(g(x)) + 1 Since we can substitute this in for g(x): Now we find the inverse of fog: + Again, we get an inverse that is not a function. For g^-1 o f^-1, we use the two inverses we found back at the start and feed the inverse of f into the inverse of g as its input: From the inverse of g, +, we can see that it finds the positive and negative square roots of its input. So it will do the same to f^1, : + which matches the inverse of fog!