Method One
f(g(x)) = x
Method Two
g(f(x)) = x
So let's pick a pair of functions and try this out.
f(x) = x^2 + 1
g(x) =sqrt(x - 1)
Using Method 1
f(g(x)) = (g(x)^2 + 1 You put a g(x) wherever you see an x in f(x)
f(g(x)) = [sqrt(x - 1)}^2 + 1 Substitute the right side of g(x) on the right side of f(x)
f(g(x)) = x - 1 + 1 Expand and cancel
f(g(x) = x
Using Method 2
g(f(x)) = sqrt(f(x) - 1) Put an f(x) wherever you see an x in g(x)
g(f(x)) = sqrt(x^2 + 1 - 1) Substitute the value of f(x) in the g(x) equation
g(f(x)) = sqrt(x^2) The 1s cancel. Take the square root of x^2
g(f(x)) = x You get x which is what you need to get.
So these two functions are the inverses of each other. Both methods confirm the results. A graph may help you to understand.
Notice how the red line (f(x) = x^2 + 1) is reflected across the green line to become the blue line (g(x) = sqrt(x - 1) ) That is another way to tell that 2 equations are inverses.
Note further that I have take the equations so that x in all three cases is ≥ 0
