Simple one...still pls explain

In place of X you simply substitute (1-X) and see for which of the functions you get the same value that you started off with.

A must do question since the choice doesnt have "None of these" thereby letting you know that one of the options must be the correct answer.

Neo2000 wrote:In place of X you simply substitute (1-X) and see for which of the functions you get the same value that you started off with.

A must do question since the choice doesnt have "None of these" thereby letting you know that one of the options must be the correct answer.

I'm having a hard time with these problems... can you please show me an example? THANKS!

First, you have to understand what f(1-x) does. As Neo2000 stated, it means every time you see an 'x', substitute it with '(1-x)'.

For example,
f(x) = 1-x

Then
f(1-x) = 1-(1-x) = 1-1+x = x


For this problem, you need find the solution where f(x) and f(1-x) are the same. If you test each answer choice, you will see that D matches that criteria.

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

f(1-x) = (1-x)^2 * [1-(1-x)]^2
(1-x)^2 * [1-1+x)]^2
(1-x)^2 * x^2
x^2 * (1-x)^2
which is f(x)

tmmyc wrote:First, you have to understand what f(1-x) does. As Neo2000 stated, it means every time you see an 'x', substitute it with '(1-x)'.

For example,
f(x) = 1-x

Then
f(1-x) = 1-(1-x) = 1-1+x = x


For this problem, you need find the solution where f(x) and f(1-x) are the same. If you test each answer choice, you will see that D matches that criteria.

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

f(1-x) = (1-x)^2 * [1-(1-x)]^2
(1-x)^2 * [1-1+x)]^2
(1-x)^2 * x^2
x^2 * (1-x)^2
which is f(x)

I FINALLY GET IT! Thank YOU!