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

I like this strategy, but I would probably pick y = 0 first, eliminate some answers quickly, then plug in y=2 for the last bit of elimination until only 1 answer remains.

Hi I understood how you re-wrote the function and ended up with - 4y^2 * 4(1-y)^2

But I do not understand how you ended up with f(y) = f(1-y)?

Could you please explain this?

We can let y = 1, and we are looking for a function in the given choices that makes f(1) = f(0).

A) f(y) = 3y + 2

f(1) = 5 and f(0) = 2

Choice A is not correct.

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

f(1) = 0 and f(0) = 1

Choice B is not correct.

C) f(y) = 4y^2 * (2-2y)^2

f(1) = 0 and f(0) = 0

Choice C could be correct, but we have to make sure that the last two choices cannot be correct so that C is the only correct answer.

(D) f(y) = y/(y-1)

f(1) is undefined; therefore we cannot reach any conclusions for this function; it could be correct.

(E) f(y) = y/(y+1)

f(1) = 1/2 and f(0) = 0

Choice E is not correct.

We have eliminated every answer choice except C and D. To decide between C and D, let’s take y = 2. The function we are looking for should satisfy f(2) = f(1 - 2) = f(-1).

(C) f(y) = 4y^2 * (2-2y)^2

f(2) = 16 * 4 and f(-1) = 4 * 16

Choice C could be correct.

(D) f(y) = y/(y - 1)

f(2) = 2 and f(-1) = 1/2

Choice D is not correct.

Answer: C

\(f(y)=4(1-y)^2(2-2(1-y))^2\)
\(f(y)=4(1-y)^2(2y)^2\)
\(f(y)=4(1-y)^24y^2\)

\(f(y)=4y^2(2-2y)^2\)
\(f(y)=4y^2(2(1-y))^2\)
\(f(y)=4y^24(1-y)^2\)

Answer is C

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