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

[dschaaf]

Can anyone provide an explanation on why the answer for "For which of the following functions is f(x) = f(1-x) for all x?" is D?

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)

The correct answer is d: f(x) = x^2 * (1-x)^2

Any help would be appreciated. Thanks.
-Doug

[cfaking]

just replace x with 1-x
you get f(1-x)

only D will return f(x)=f(1-x)


regds
Nitya

[Ben Ku]

When you have function questions, just plug in for x whatever is in the parentheses. For example, if f(x) = x^2 - 2x + 1, and you're finding f(2), then plug in 2 everywhere you see x: (2)^2 - 2(x) + 1.

It's the same thing here. You're given f(x); to find f(1-x), plug into each function "1 - x" wherever you see "x".

(A) f(x) = 1-x
so f(1-x) = 1 - (1-x) = x
(B) f(x) = 1-x^2
so f(1-x) = 1 - (1-x)^2 = 1 - (1 - 2x + x^2) = 2x - x^2
(C) f(x) = x^2 - (1-x)^2
so f(1-x) = (1-x)^2 - (1 - (1-x))^2 = 1 - 2x + x^2 - (x)^2 = 1 - 2x
(D) f(x) = x^2 * (1-x)^2
so f(1-x) = (1 - x)^2 * (1 - (1 - x))^2 = (1 - x)^2 * (x)^2
(E) f(x) = x/(1-x)
so f(1-x) = (1-x)/(1-(1-x)) = (1-x)/x

Hope that helps.

[khushbumerchant]

When you have function questions, just plug in for x whatever is in the parentheses. For example, if f(x) = x^2 - 2x + 1, and you're finding f(2), then plug in 2 everywhere you see x: (2)^2 - 2(x) + 1.

It's the same thing here. You're given f(x); to find f(1-x), plug into each function "1 - x" wherever you see "x".

(A) f(x) = 1-x
so f(1-x) = 1 - (1-x) = x
(B) f(x) = 1-x^2
so f(1-x) = 1 - (1-x)^2 = 1 - (1 - 2x + x^2) = 2x - x^2
(C) f(x) = x^2 - (1-x)^2
so f(1-x) = (1-x)^2 - (1 - (1-x))^2 = 1 - 2x + x^2 - (x)^2 = 1 - 2x
(D) f(x) = x^2 * (1-x)^2
so f(1-x) = (1 - x)^2 * (1 - (1 - x))^2 = (1 - x)^2 * (x)^2
(E) f(x) = x/(1-x)
so f(1-x) = (1-x)/(1-(1-x)) = (1-x)/x

Hope that helps.

Ok. I got baffled in this Q during my test. Here's what I did:

D. f(x) = x^2 *(1-x)^2
So, f(1-x) = (1-x)^2 * [1-(1-x)]^2
= (1-2x - x^2) * [x]^2
= x^2 - 2X^3 - X^4

And i believe this is no where, close to the function. So where did I go wrong here, can anybody help me out?

[jnelson0612]

When you have function questions, just plug in for x whatever is in the parentheses. For example, if f(x) = x^2 - 2x + 1, and you're finding f(2), then plug in 2 everywhere you see x: (2)^2 - 2(x) + 1.

It's the same thing here. You're given f(x); to find f(1-x), plug into each function "1 - x" wherever you see "x".

(A) f(x) = 1-x
so f(1-x) = 1 - (1-x) = x
(B) f(x) = 1-x^2
so f(1-x) = 1 - (1-x)^2 = 1 - (1 - 2x + x^2) = 2x - x^2
(C) f(x) = x^2 - (1-x)^2
so f(1-x) = (1-x)^2 - (1 - (1-x))^2 = 1 - 2x + x^2 - (x)^2 = 1 - 2x
(D) f(x) = x^2 * (1-x)^2
so f(1-x) = (1 - x)^2 * (1 - (1 - x))^2 = (1 - x)^2 * (x)^2
(E) f(x) = x/(1-x)
so f(1-x) = (1-x)/(1-(1-x)) = (1-x)/x

Hope that helps.

Ok. I got baffled in this Q during my test. Here's what I did:

D. f(x) = x^2 *(1-x)^2
So, f(1-x) = (1-x)^2 * [1-(1-x)]^2
= (1-2x - x^2) * [x]^2
= x^2 - 2X^3 - X^4

And i believe this is no where, close to the function. So where did I go wrong here, can anybody help me out?

You've actually done just great! On that last line, try to first factor out an x^2. Then see if you can rearrange to make it look like your first line.