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

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

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20 Nov 2005, 05:33
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For which of the following functions f is f(x) = f(1-x) for all 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)

OPEN DISCUSSION OF THIS QUESTION IS HERE: for-which-of-the-following-functions-f-is-f-x-f-1-x-for-85751.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Sep 2014, 22:48, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]

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20 Nov 2005, 06:42
Should be D. F(x)=(x^2) (1-x^2)

D. is the only function which is the greatest at 1/2.
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]

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20 Nov 2005, 10:12
njss750 wrote:
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

I remember seeing this problem in a GMATprep test. I think choice D should be F(x)=(x^2) (1-x)^2. If that is true then substituting (1-x) for will give us the same function back and the answer choice is D, where the function is multiplicative
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]

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25 Nov 2005, 07:09
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]

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25 Sep 2014, 22:06
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]

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25 Sep 2014, 22:49
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njss750 wrote:
For which of the following functions f is f(x) = f(1-x) for all 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)

$$f(x)="some \ expression \ with \ variable \ x"$$, means that the value of $$f(x)$$ can be found by calculating the expression for the particular $$x$$.

For example: if $$f(x)=3x+2$$, what is the value of $$f(3)$$? Just plug $$3$$ for $$x$$, $$f(3)=3*3+2=11$$, so if the function is $$f(x)=3x+2$$, then $$f(3)=11$$.

There are some functions for which $$f(x)=f(-x)$$. For example: if we define $$f(x)$$ as $$f(x)=3*x^2+2$$, the value of $$f(x)$$ will be always positive and will give the following values: for $$x=-5$$, $$f(x)=3*(-5)^2+2=77$$; for $$x=0$$, $$f(0)=3*0^2+2=2$$. Please note that $$f(x)$$ in this case is equal to $$f(-x)$$, meaning that for positive values of $$x$$ you'll get the same values of $$f(x)$$ as for the negative values of $$x$$.

So, basically in original question we are told to define the expression, for which $$f(x)=f(1-x)$$, which means that plugging $$x$$ and $$1-x$$ in the expression must give same result.

A. $$f(x)=1-x$$ --> $$1-x$$ is the expression for $$f(x)$$, we want to find whether the expression for $$f(1-x)$$ would be the same: plug $$1-x$$ --> $$f(1-x)=1-(1-x)=x$$. As $$1-x$$ and $$x$$ are different, so $$f(x)$$ does not equal to $$f(1-x)$$.

The same with the other options:

(A) $$f(x)=1-x$$, so $$f(1-x)=1-(1-x)=x$$ --> $$1-x$$ and $$x$$: no match.

(B) $$f(x)=1-x^2$$, so $$f(1-x)=1-(1-x)^2=1-1+2x-x^2=2x-x^2$$ --> $$1-x^2$$ and $$2x-x^2$$: no match.

(C) $$f(x)=x^2-(1-x)^2=x^2-1+2x+x^2=2x-1$$, so $$f(1-x)=2(1-x)-1=1-2x$$ --> $$2x-1$$ and $$1-2x$$: no match.

(D) $$f(x)=x^2*(1-x)^2$$, so $$f(1-x)=(1-x)^2*(1-1+x)^2=(1-x)^2*x^2$$ --> $$x^2*(1-x)^2$$ and $$(1-x)^2*x^2$$. Bingo! if $$f(x)=x^2*(1-x)^2$$ then $$f(1-x)$$ also equals to $$x^2*(1-x)^2$$.

Still let's check (E)

(E) $$f(x)=\frac{x}{1-x}$$ --> $$f(1-x)=\frac{1-x}{1-1+x}=\frac{1-x}{x}$$. $$\frac{x}{1-x}$$ and $$\frac{1-x}{x}$$: no match.

But this problem can be solved by simple number picking: plug in numbers.

As stem says that "following functions f is f(x) = f (1-x) for all x", so it should work for all choices of $$x$$.

Now let $$x$$ be 2 (note that: -1, 0, and 1 generally are not good choices for number picking), then $$1-x=1-2=-1$$. So we should check whether $$f(2)=f(-1)$$.

(A) $$f(2)=1-x=1-2=-1$$ and $$f(-1)=1-(-1)=2$$ --> $$-1\neq{2}$$;

(B) $$f(2)=1-x^2=1-4=-3$$ and $$f(-1)=1-1=0$$ --> $$-3\neq{0}$$;

(C) $$f(2)=x^2-(1-x)^2=x^2-1+2x+x^2=2x-1=2*2-1=3$$ and $$f(-1)=2*(-1)-1=-3$$ --> $$3\neq{-3}$$;

(D) $$f(2)=x^2*(1-x)^2=(-2)^2*(-1)^2=4$$ and $$f(-1)=(-1)^2*2^2=4$$ --> $$4=4$$, correct;

(E) $$f(2)=\frac{x}{1-x}=\frac{2}{1-2}=-2$$ and $$f(-1)=\frac{-1}{1-(-1)}=-\frac{1}{2}$$ --> $$-2\neq{-\frac{1}{2}}$$.

It might happen that for some choices of $$x$$ other options may be "correct" as well. If this happens, just pick some other number for $$x$$ and check again these "correct" options only.

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OPEN DISCUSSION OF THIS QUESTION IS HERE: for-which-of-the-following-functions-f-is-f-x-f-1-x-for-85751.html
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Re: For which of the following functions f is f(x) = f(1-x) for all x?   [#permalink] 25 Sep 2014, 22:49
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