make a response to this:

Question

. f(x) = 2x + 3 2. g(x) = x - 1 For f(x) = 2x + 3: To find the inverse, we swap x and y and solve for y: x = 2y + 3 x - 3 = 2y y = (x - 3)/2 So, the inverse of f(x) is f?¹(x) = (x - 3)/2. For g(x) = x - 1: To find the inverse, we swap x and y and solve for y: x = y - 1 y = x + 1 So, the inverse of g(x) is g?¹(x) = x + 1. Consider the composite function h(x) = f(g(x)): h(x) = f(g(x)) = f(x - 1) = 2(x - 1) + 3 = 2x - 2 + 3 = 2x + 1 To find the inverse, we swap x and y and solve for y: x = 2y + 1 x - 1 = 2y y = (x - 1)/2 So, the inverse of h(x) is h?¹(x) = (x - 1)/2. To verify if the inverse of the composite function is a function. In this case, h?¹(x) = (x - 1)/2 is a function because it passes the vertical line test (each input x maps to exactly one output y). Therefore, if the inverses of two functions are both functions, the inverse of the composite function made by the original functions will also be a function.

Answer 1

The conclusion may be true, but simply giving an example that is true does not prove the proposition.

I believe you must also show that the domain and ranges are identical sets. In function terminology, the component functions must be both one-to-one and onto.