For which of the following functions f is f(x) = f(1-x) for all x?

f(x) = 1-x
f(x) = 1-x^2
f(x) = x^2 - (1-x)^2
f(x) = x^2(1-x)^2
f(x) = x/1-x

For which of the following functions f is f(x) = f(1-x) for all x?

f(x) = 1-x
f(x) = 1-x^2
f(x) = x^2 - (1-x)^2
f(x) = x^2(1-x)^2
f(x) = x/1-x

Hi, You need to substitute x & 1-X in each of the functions to find out if LHS = RHS.

(1) f(x) = 1-x -> LHS
f(1-x) = 1-(1-x) = x -> RHS
so, LHS <> RHS

(2) f(x) = 1-x^2 -> LHS
f(1-x) = 1- (1-x)^2 = 1 - 1 - x^2 + 2*x = 2x - x^2 -> RHS
so, LHS <> RHS

(3)f(x) = x^2 - (1-x)^2 = -1-2x
f(1-x) = (1-x)^2 - (1- (1-x))^2 = 1+x^2-2x - (1-1-x^2+2x) = 1+2x^2-4x
so, again LHS <> RHS

(4)f(x) = x^2(1-x)^2
f(1-x) = (1-x)^2 * (1-(1-x))^2 = (1-x)^2 * x^2.
LHS = RHS. HEnce this is the answer.

(5) f(x) = x/1-x
f(1-x) = 1-x/x
LHS <> RHS

Quite a lengthy problem :) but got to be very careful with signs and simplification.

What is this LHS business????

LHS refers to Left Hand Side & RHS refers to Right Hand Side. Sorry for the delayed response.

ok, yeah, the last solution is correct. in fact, i owe a beer to whoever posted it, because i certainly don't have the patience to type in all that algebra.

BUT

like omg!
that's a lot of work for two minutes.

this problem says for ALL x. this means that you can plug in any value of x you want, carry it through, and just check whether it works.
if an answer choice works, it's not definitively the correct answer (because it may have just been a coincidence) - but, if an answer choice is ever wrong, you can cross it out.

let's try plugging in 0 for x
the question becomes:
for which of the following functions is f(0) = f(1)?
a: f(0) = 1, f(1) = 0. nope
b: f(0) = 1, f(1) = 0. nope
c: f(0) = -1, f(1) = 1. nope
d: f(0) = 0, f(1) = 0. looks good
e: f(0) = 0, f(1) = undefined. nope

d wins.

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LHS and RHS were the two rival high schools in my hometown.
;)