Find start and end of week and day in SQL Server

Yet another attempt to build up a library of SQL Server functions. As I found out, not all my previous functions did work properly in all environments.

-- finds start of current week for the provided date
CREATE FUNCTION [dbo].[fnStartOfWeek]
(
  @INPUTDATE DATETIME
)
RETURNS DATETIME

AS
BEGIN
  -- THIS does not work in function.
  -- SET DATEFIRST 1 -- set monday to be the first day of week.

  DECLARE @DOW INT -- to store day of week
  SET @DOW = DATEPART(DW, @INPUTDATE)

  -- Magic convertion of monday to 1, tuesday to 2, etc.
  -- irrespect what SQL server thinks about start of the week.
  -- But here we have sunday marked as 0, but we fix this later.
  SET @DOW = (@DOW + @@DATEFIRST - 1) %7
  IF @DOW = 0 SET @DOW = 7 -- fix for sunday

  RETURN dbo.fnStartOfDay( DATEADD(DD, 1 - @DOW,@INPUTDATE) )

END

GO


-- finds end of the current week for the provided date
CREATE FUNCTION [fnEndOfWeek]
(
	@INPUTDATE DATETIME
)
RETURNS DATETIME
AS
BEGIN
	DECLARE @WEEKSTART DATeTIME,
			@WEEKEND DATETIME
	SET @WEEKSTART = dbo.fnStartOfWeek(@INPUTDATE) -- Find start of this week.
	SET @WEEKEND = DATEADD(week,1, @WEEKSTART)		-- add one week to the start of the week.
	SET @WEEKEND = DATEADD(second, -1, @WEEKEND)	-- take out one second from the end of the week
	RETURN @WEEKEND
END

GO 

-- Finds end of the day for the provided date
CREATE FUNCTION [fnEndOfDay]
(
	@DATE DATETIME
)
RETURNS DATETIME
AS
BEGIN
	RETURN DATEADD(s,-1,DATEADD(d, DATEDIFF(d,0,@DATE)+1,0))
END

GO

-- Finds Start of the day for the provided day
CREATE FUNCTION [fnStartOfDay]
(
	@DATE DATETIME
)
RETURNS DATETIME
AS
BEGIN
	Return DATEADD(d, DATEDIFF(d,0,@DATE),0)
END