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

Question

(A) f(y) = 3y + 2
(B) f(y) = 1-y^2
(C) f(y) = 4y^2 * (2-2y)^2
(D) f(y) = y/(y-1)
(E) f(y) = y/(y+1)

FacelessMan · Answer

4y^2 * (2-2y)^2 can be written as 4y^2 * 4(1-y)^2. Now for this f(y) = f(1-y). Hence C.

fantaisie · Answer

I just picked a random number and tested it:

y = 2
 
a) f(2) = 6*2 + 2 = 14; f(-1) = -3 + 2 = -1 NO
b) f(2) =1 - 2^2 = 1 - 4 = -3 ; f(-1) = 1 - (-1)^2 = 1 - 1 = 0 NO
c) f(2) = 4*2^2 * (2-2*2)^2 = 4*4 * (2-4)^2 = 16 * 4; f(-1) = 4*1 * (2-2(-1))^2 = 4 * (2+2)*2 = 4 * 16 YES

Here we could stop and safely pick the number without doing d & e, but just in case:

d) f(2) = 2/1 = 2 ; f(-1) = -1 / -2 = 1/2 NO
e) f(2) = 2/3; f(-1) = -1/0 = 0 (div by 0) NO

So I proved that in all but C case if y=2 the values of f(y) ≠ f(1-y)

Answer:  C

pourts · Answer

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.

Gradus · Answer

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?

ScottTargetTestPrep · Answer

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

BlueRocketAsh · Answer

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

bumpbot · Answer

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

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

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