(g o f)(x) is the same as g(f(x))
f(x) = x+1 g(x) = x² - x
g(f(x)) = g(x+1)
Note g(x) = x² - x, therefore g(x+1) would be that anywhere we see x in g(x) we replace it with (x+1)
g(x) = x² - x
g(x+1) = (x+1)² - (x+1) = x² + 2x +1 - x - 1 = x² + 2x-x +1 -1 = x² + x
Therefore (g o f)(x) = x² + x
