A function is a type of black box: you put a value in the box, and after processing, another value is output. The formula for the function gives instructions for processing/transforming the value that is put inside the box.

For example, f(x) =3x tells us that when x is put inside the box, after processing the result will be 3x. For instance, if 4 is put inside the box, the output will be 3*4. So f(4)=3*4.

To process the input, just replace x wherever you see it, with the actual value (or variable) that is put inside the box. For instance f(2y) means that we are putting 2y into the black box, and we will replace x wherever we see it with 2y. Since f(x)=3x in our example function, f(2y)=3*(2y).

In this question, we are asked "For which function is f(x)=f(1-x)?" This means that we need to find a function such that if we put 'x' into the box, the output after processing will be the same as if we put '1-x' into the box of that same function.

The answer choices already tell us what f(x) is (they tell us what the output is when x is put into the box). So all we need to do is find out what f(1-x) is (what the output is when 1-x is put into the box).

D is the right answer because if we put 1-x into the that function (remember that to process the input just replace x wherever you see it with the input value, in this case 1-x) we get:

D. f(1-x) = (1-x)^2 [1-(1-x)]^2

After simplifying this, you should find that the output (the right side of the equation above) is identical to the output when x is put into the box (the right side of the original answer choice D)

This was just an overview of the logic. 2 detailed explanations (plug-in and algebra) as well as a video solution are available at GMATPrep Question 1142. If you struggle with this type of problem, use the drill engine to generate timed drills and set topic='Functions & Sequences'

Hope this helps,
-Patrick

This question is asking which formula yields the same result for f(x) and f(x-1).

The quickest way to do this is to plug in numbers instead of messing around with the algebra.

So which numbers would work best?

Let x = 1
Therefore x-1 = 0
So now just evaluate all the formulas for x=1 and x=0. The formula that gives you the same answer for both is the right one.

Hi,
I still find this problem hard to solve..Can an expert please tell me what is it that I am doing wrong? I know it is in the first step, so I will only write the first step cuz I'll know how to tackle afterwards-Thank you!

f(x)= f(1-x),then
x^2 (1-x)^2 = (1-[x^2(1-x)^2]

after solving it, I still have the 1 that's before the -x remaining..

It took me forever to understand this question but I believe that I have finally gotten it right. And, now that I have understood it looks pretty simple to me.

The question says:
f(x) = f(1-x)

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

Which basically means that for the solution of the right side of the equation (for any given option) should be the same as when X is replaced with "1-x". Since f(x)=f(1-x).

Now lets take the first option:

A)f(x) = 1-x

We need to get the same solution for f(x) = 1-x and 1-(1-x)

If we replace X with 3 in f(x) = 1-x and 1-(1-x), we get -2 and 4 respectively and therefore, we can move on to the next option.

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

We need to get the same solution for f(x) = 1-x^2 and 1-(1-x)^2

If we replace X with 3 in f(x) = 1-x^2 and 1-(1-x)^2, we get -8 and -3 respectively and therefore, we can move on to the next option.

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

We need to get the same solution for f(x) = x^2 - (1-x)^2 and (1-x)^2 -(1-(1-x))^2

If we replace X with 3 in f(x)= x^2 -(1-x)^2 and (1-x)^2 -(1-(1-x))^2, we get 5 and -5 respectively and therefore, we can move on to the next option.

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

We need to get the same solution for f(x) = x^2(1-x)^2 and (1-x)^2(1-(1-x))^2

If we replace X with 3 in f(x)= x^2(1-x)^2 and(1-x)^2(1-(1-x))^2, we get 36 and 36 respectively. This is it! Its satisfying the requirement of f(x) = f(1-x) for all values.

Please let me know if my approach is wrong.

Thanks