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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