My Math Forum Demonstration (symmetry)

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 September 7th, 2009, 11:23 AM #1 Newbie   Joined: Jul 2009 Posts: 11 Thanks: 0 Demonstration (symmetry) Demonstrate that if x=a axis of symmetry for f(x) then: $f(x) = f(2a - x) \$ x-from R.
 September 7th, 2009, 11:59 AM #2 Member   Joined: Jul 2009 Posts: 34 Thanks: 0 Re: Demonstration (symmetry) If x and y have the same distance from a and $y>x$, then $a-x=y-a$, and then $f(x)=f(y)$ from $a-x=y-a$ we get $y=2a-x$, so we get $f(x)=f(2a-x)$ In the same way, if $y, then $a-y=x-a$, and then too $f(y)=f(x)$ from $a-y=x-a$ we get again $y=2a-x$, so we get $f(x)=f(2a-x)$ The last option that $y=x$ we have $y=x=a$, and then it's obvious that $f(x)=f(2a-x)$
 September 9th, 2009, 09:15 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,927 Thanks: 2205 Too long-winded! By definition of symmetry, f(a - t) = f(a + t) for all real values of t. Putting t = a - x gives f(x) = f(2a - x) for all real values of x.

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