Answer and step-by-step explanation:
We just have to apply one function in the other one. See:
→ f(g(x)):
[tex]f(x) = \dfrac{2x-3}{x+4}\\\\f(g(x)) = \dfrac{2g(x)-3}{g(x)+4} = \dfrac{2\cdot\dfrac{3+4x}{2-x}-3}{\dfrac{3+4x}{2-x}+4}\\\\f(g(x)) = \dfrac{\dfrac{6+8x}{2-x}-3}{\dfrac{3+4x}{2-x}+4}\\\\f(g(x)) = \dfrac{\dfrac{(6+8x)-3(2-x)}{2-x}}{\,\,\,\dfrac{(3+4x)+4(2-x)}{2-x}\,\,\,}=\dfrac{(6+8x)-3(2-x)}{(3+4x)+4(2-x)}\\\\f(g(x)) =\dfrac{6+8x-6+3x}{3+4x+8-4x}=\dfrac{11x}{11}\\\\\\\boxed{f(g(x)) = x}[/tex]
→ g(f(x)):
[tex]g(x) = \dfrac{3+4x}{2-x}\\\\g(f(x)) = \dfrac{3+4f(x)}{2-f(x)} = \dfrac{3+4\dfrac{2x-3}{x+4}}{2-\dfrac{2x-3}{x+4}}\\\\g(f(x)) = \dfrac{3+\dfrac{8x-12}{x+4}}{2-\dfrac{2x-3}{x+4}}\\\\g(f(x)) = \dfrac{\dfrac{3(x+4)+(8x-12)}{x+4}}{\,\,\,\dfrac{2(x+4)-(2x-3)}{x+4}\,\,\,}=\dfrac{3(x+4)+(8x-12)}{2(x+4)-(2x-3)}\\\\g(f(x)) =\dfrac{3x+12+8x-12}{2x+8-2x+3}=\dfrac{11x}{11}\\\\\\\boxed{g(f(x)) = x}[/tex]
What verifies that f and g are inverses of each other.
Q.E.D.
