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

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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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Okay, so you need to check if f(x) = f(1-x)

Let's do this one by one.

Case A: - Wrong
$$f(x) = 1-x$$
$$f(1-x) = 1-(1-x) = x$$

So we clearly see that f(x) is not f(1-x)

Case B: Wrong

$$f(x) = 1-x-x^2$$
$$f(1-x) = 1-(1-x)-(1-x)^2 = x - 1 - x^2 + 2x = 3x-1-x^2$$

Case C: Wrong

$$f(x) = x^2 - (1-x)^2 = x^2 - 1 +2x - x^2= 2x-1$$
$$f(1-x) = (1-x)^2 - (1-(1-x))^2 = (1-x)^2 - (x)^2= 1-2x$$

Case D: Right

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

Let's check E just to be sure.

Case E: Wrong

$$f(x) = \frac{x}{1-x}$$
$$f(1-x) = \frac{1-x}{1-(1-x)}=\frac{1-x}{x}$$
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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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Bunuel:

sometimes, I almost die from the awesomeness of your posts.
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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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This one is quite simple. We have to substitute (1 - x) for x in each of the options and see which option yields the same result for both f(x) and f(1 - x).

However, careful observation of the options will easily tell you that only D will fulfill this condition because in D, the x now becomes (1 - x) and (1 - x) now becomes x. The overall outcome will remain the same. You don't even need to expand or do any calculations whatsoever.
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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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You need to check for every option.
Either you can replace x in every option with 1-x or you can take x = 1 and then check for which option f(1) = f(1-1) = f(0).
In A f(1) = 1-1 = 0. f(0) = 1-0 = 1.
In B f(1) = 1 - 1^2 = 0. f(0) = 1 - 0 = 1
In C f(1) = 1^2 - (1 - 1 )^2 = 1. f(0) = 0 - (1-0)^2 = 0 - 1 = -1
in D f(1) = 1(1-1)^2 = 0. f(0) = 0.(1-0)^2 = 0. ( Right answer )
And for option E you can take x = 2 because if you take x = 1 in denominator the denominator becomes zero, which makes the f(x) undefined.
f(2) = 2/(1 - 2) = -2. f(1-2) = f(-1) = -1/(1-(-1)) = -1/2.

Please give a kudo if you like my explanation.
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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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study 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)

Since we are not given any restrictions on the value of x, let’s let x = 1. Thus, we are determining for which of the following functions is f(1) = f(1-1), i.e., f(1) = f(0). Next, we can test each answer choice using our value x = 1.

A. f(x) = 1 - x

f(1) = 1 - 1 = 0

f(0) = 1 - 0 = 1

Since 0 does not equal 1, A is not correct.

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

f(1) = 1 - 1^2 = 1 - 1 = 0

f(0) = 1 - 0^2 = 1 - 0 = 1

Since 0 does not equal 1, B is not correct.

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

f(1) = 1^2 - (1 - 1)^2 = 1 - 0 = 1

f(0) = 0^2 - (1 - 0)^2 = 0 - 1 = -1

Since 1 does not equal -1, C is not correct.

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

f(1) = 1^2*(1 - 1)^2 = 1(0)= 0

f(0) = 0^2*(1 - 0)^2 = 0(2) = 0

Since 0 equals 0, D is correct.

Alternate Solution:

Let’s test each answer choice using x and 1 - x.

A. f(x) = 1 - x

f(x) = 1 - x

f(1 - x) = 1 - (1 - x) = x

Since 1 - x does not equal x, A is not correct.

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

f(x) = 1 - x^2

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

Since 1 - x^2 does not equal 2x - x^2, B is not correct.

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

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

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

Since 2x - 1 does not equal 1 - 2x, C is not correct.

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

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

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

Since x^2*(1 - x)^2 equals (1 - x)^2*x^2, D is correct.

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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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study 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. $$f(x) = \frac{x}{(1 - x)}$$

Let's try plugging in an easy value for x. How about x = 0.
So, we can reword the question as, For which of the following functions is f(0)=f(1-0)
In other words, we're looking for a function such that f(0) = f(1)

A) f(x)=1-x
f(0)=1-0 = 1
f(1)=1-1 = 0
Since f(0) doesn't equal f(1), eliminate A

B) f(x) = 1 - x^2
f(0) = 1 - 0^2 = 1
f(1) = 1 - 1^2 = 0
Since f(0) doesn't equal f(1), eliminate B

C) f(x) = x^2 - (1-x)^2
f(0) = 0^2 - (1-0)^2 = -1
f(1) = 1^2 - (1-1)^2 = 1
Since f(0) doesn't equal f(1), eliminate C

D) f(x) = x^2(1-x)^2
f(0) = 0^2(1-0)^2 = 0
f(1) = 1^2(1-1)^2 = 0
Since f(0) equals f(1), keep D for now

E) f(x) = x/(1-x)
f(0) = 0/(1-0) = 0
f(1) = 1/(1-1) = undefined
Since f(0) doesn't equal f(1), eliminate E

Since only D satisfies the condition that f(x)=f(1-x) when x=0, the correct answer is D

Cheers,
Brent
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For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
Bunuel wrote:
study 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)

I've already seen quite a few GMAT questions of this type. They are quite easy to solve, since you understand the concept.

$$f(x)="some \ expression \ with \ variable \ x"$$, 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.

Similar questions:
https://gmatclub.com/forum/for-which-of- ... 93184.html
https://gmatclub.com/forum/for-which-of- ... 24491.html
https://gmatclub.com/forum/let-the-funct ... 43311.html

Hope it helps.

Bunuel for option D

$$f(x)=x^2* (1-x)^2 = x^2 * ( 1-2x+x^2)$$ = now plugged in (1-x) for x --> $$(1-x)^2 * (1-2(1-x)+(1-x)^2)$$

got $$(1-x)^2 * (1-2+2x+1-2x+x^2)$$ obviously smth is not right abt it

what am doing wrong ?

And why in Option C you didnt plug in 1-x to get 2x-1 whereas in D you plugged in x-1 straight away without expanding (1-x)^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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dave13 wrote:

Bunuel for option D

$$f(x)=x^2* (1-x)^2 = x^2 * ( 1-2x+x^2)$$ = now plugged in (1-x) for x --> $$(1-x)^2 * (1-2(1-x)+(1-x)^2)$$

got $$(1-x)^2 * (1-2+2x+1-2x+x^2)$$ obviously smth is not right abt it

what am doing wrong ?

And why in Option C you didnt plug in 1-x to get 2x-1 whereas in D you plugged in x-1 straight away without expanding (1-x)^2 ?

In C, even if you substitute x =1-x, you will get the same answer.

In D, 1-2+2x+1-2x+x^2 and you will still get your answer as everything will get canceled except x^2
$$(1-x)^2=1+x^2-2x....1+(1-x)^2-2(1-x)=1+1+x^2-2x-2+2x=x^2$$
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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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dave13 wrote:
Bunuel wrote:
study 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)

I've already seen quite a few GMAT questions of this type. They are quite easy to solve, since you understand the concept.

$$f(x)="some \ expression \ with \ variable \ x"$$, 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.

Similar questions:
https://gmatclub.com/forum/for-which-of- ... 93184.html
https://gmatclub.com/forum/for-which-of- ... 24491.html
https://gmatclub.com/forum/let-the-funct ... 43311.html

Hope it helps.

Bunuel for option D

$$f(x)=x^2* (1-x)^2 = x^2 * ( 1-2x+x^2)$$ = now plugged in (1-x) for x --> $$(1-x)^2 * (1-2(1-x)+(1-x)^2)$$

got $$(1-x)^2 * (1-2+2x+1-2x+x^2)$$ obviously smth is not right abt it

what am doing wrong ?

And why in Option C you didn`t plug in 1-x to get 2x-1 whereas in D you plugged in x-1 straight away without expanding (1-x)^2 ?

When you replace x by (1 - x) in (1 - x), you get x back!

(1 - x) becomes (1 - (1-x)) which is (1 - 1 + x) which is x.

Hence

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

becomes (1-x)^2 - x^2

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

becomes (1-x)^2 * x^2

In (C) the expression changes but (D) doesn't because multiplication is commutative (xy = yx)
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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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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
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For which of the following functions f is f(x) = f (1-x) for all x?

Lets try plugging in values to solve this question. Since there is no condition given for x , we can assume any values for x.
Assume x= 2, then f(2) = f(1-2) = f(-1). So we need to check in which functions given below, f(2)= f(-1)?

A. f (x) = 1 - x
f(2)=1-2 =-1
f(-1) = 1+1 =2 ==> f(2) ≠ f(-1)

B. f (x) = 1 - x^2
f(2) = 1-4 = -3
f(-1) = 1-1 = 0 ==> f(2) ≠ f(-1)

C. f (x) = x^2 - (1 - x)^2
f(2)= 4-(-1)^2 = 4-1 = 3
f(-1) = 1-4 = -3 ==> f(2) ≠ f(-1)

D. f (x) = x^2 (1 - x)^2
f(2) = 4*1=4
f(-1) = 1*4 =4 ==> f(2)= f(-1)

E. f (x) = x/(1 - x)
f(2)= 2/-1 = -2
f(-1) = -1/2 ==> f(2) ≠ f(-1)

Option D is the correct answer.

Thanks,
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
ScottTargetTestPrep wrote:
study 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)

Since we are not given any restrictions on the value of x, let’s let x = 1. Thus, we are determining for which of the following functions is f(1) = f(1-1), i.e., f(1) = f(0). Next, we can test each answer choice using our value x = 1.

A. f(x) = 1 - x

f(1) = 1 - 1 = 0

f(0) = 1 - 0 = 1

Since 0 does not equal 1, A is not correct.

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

f(1) = 1 - 1^2 = 1 - 1 = 0

f(0) = 1 - 0^2 = 1 - 0 = 1

Since 0 does not equal 1, B is not correct.

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

f(1) = 1^2 - (1 - 1)^2 = 1 - 0 = 1

f(0) = 0^2 - (1 - 0)^2 = 0 - 1 = -1

Since 1 does not equal -1, C is not correct.

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

f(1) = 1^2*(1 - 1)^2 = 1(0)= 0

f(0) = 0^2*(1 - 0)^2 = 0(2) = 0

Since 0 equals 0, D is correct.

Alternate Solution:

Let’s test each answer choice using x and 1 - x.

A. f(x) = 1 - x

f(x) = 1 - x

f(1 - x) = 1 - (1 - x) = x

Since 1 - x does not equal x, A is not correct.

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

f(x) = 1 - x^2

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

Since 1 - x^2 does not equal 2x - x^2, B is not correct.

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

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

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

Since 2x - 1 does not equal 1 - 2x, C is not correct.

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

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

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

Since x^2*(1 - x)^2 equals (1 - x)^2*x^2, D is correct.

Hi Bunuel and ScottTargetTestPrep,
Can we infer that if the function question stem doesn't specify any restrictions on the value of a given variable, we can pluging any number to test the function?
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
KarishmaB wrote:
metallicafan wrote:
Is there a faster way to solve this question rather than replacing each "x" by (1-x)? Thanks!

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. $$f(x) = x / (1-x)$$

Tip: Try to first focus on the options where terms are added/multiplied rather than subtracted/divided. They are more symmetrical and a substitution may not change the expression. I will intuitively check D first since it involves multiplication of the terms.

Thank you KarishmaB for your amazing tip!
­Is this tip is suggested for all exercises involving algebra and number plugging?
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Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
2
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Quote:
KarishmaB metallicafanIs there a faster way to solve this question rather than replacing each "x" by (1-x)? Thanks!

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. $$f(x) = x / (1-x)$$Tip: Try to first focus on the options where terms are added/multiplied rather than subtracted/divided. They are more symmetrical and a substitution may not change the expression. I will intuitively check D first since it involves multiplication of the terms.Thank you KarishmaB for your amazing tip!
­Is this tip is suggested for all exercises involving algebra and number plugging?

The tip is useful when we are looking for symmetry. e.g. $$f(x) = f(1-x)$$ for all x?

I know that a + b = b + a but (a - b) is not equal to (b - a).
Similarly, a * b = b * a but ­a/b is not equal to b/a.
Hence addition and multiplications are more likely to give me same result when the terms are exchanged.

It could be applicable in any topic depending on context.
Re: For which of the following functions f is f(x) = f(1-x) for all x? [#permalink]
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