Prelude> let f x = x * 2
Prelude> :t f
f :: Num a => a -> a

The typeclass Num is defined like this:

class  Num a  where
    (+), (-), (*)       :: a -> a -> a
    ...

...which means that the operator (*) expects two parameters of type a and returns a value of type a.

Since the definition of expr looks like this:
Prelude> let f x = x * 2

...and 2 is an Int, I would expect that the type inferred for (*)  is (Int -> Int -> Int) and thus f should be (Int -> Int)

Can somebody explain this?