Answer:
See proof below.
Step-by-step explanation:
Note:
I think you mean that f(x) = (x + 3)/2, which is [tex] f(x) = \dfrac{x + 3}{2} [/tex] and not what you wrote which means [tex] f(x) = x + \dfrac{3}{2} [/tex]
To prove that functions f(x) and g(x) are inverses of each other, you must do the composition of functions f and g, and then the composition of functions g and f. If both compositions give you the result of just x, then the functions are inverses of each other.
[tex] f(x) = \dfrac{x + 3}{2}; g(x) = 2x - 3 [/tex]
[tex] (f \circ g)(x) = f(g(x)) = \dfrac{g(x) + 3}{2} = \dfrac{2x - 3 + 3}{2} = \dfrac{2x}{2} = x [/tex]
[tex] (g \circ f)(x) = g(f(x)) = 2(f(x)) - 3 = \dfrac{2(x + 3)}{2} - 3 = x + 3 - 3 = x [/tex]
Since both compositions result in x, functions f(x) and g(x) are proved to be inverses of each other.
