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It means function composition. See this question.
Note also the f.g.h x is not equivalent to (f.g.h) x, because it is interpreted as f.g.(h x) which won't typecheck unless (h x) returns a function.
This is where the $ operator can come in handy: f.g.h $ x turns x from being a parameter to h to being a parameter to the whole expression. And so it becomes equivalent to f(g(h x)) and the pipe works again.