<s>[INST] For which of the following functions $f$ is $f(x)=f(2-x)$ for all $x$? The options are:
$A)$ $f(x) = x(x+2)$
$B)$ $f(x) = x-2$
$C)$ $f(x) = 2-x$
$D)$ $f(x) = 3x(x-2)$
$E)$ $f(x) = x^2(2-x)^2$
The answer given in my worksheet is $E)$ and I got it. My doubt is why it can't be $D)$ also? Solving thus:
$f(x) = 3x(x-2)$
$f(2-x) = 3(2-x)(2-x-2) = 3(2-x)(-x) = 3[-(2-x)](x) = 3(x-2)x = 3x(x-2) = f(x)$
If I substitute some random values for $x$, then it works as well.
Example $1$: $x = 3 \rightarrow f(x) = f(3) = 3(3)(3-2) = 9$
$f(2-x) = f(2-3) = f(-1) = 3(-1)(-1-2) = 9$
Example $2$: $x = -1/2 \rightarrow f(x) = f(-1/2) = 3(-1/2)(-1/2-2) = 15/4$
$f(2-x) = f(2-(-1/2)) = f(5/2) = 3(5/2)(5/2-2) = 3(5/2)(1/2) = 15/4$
Thus, I think $D)$ should be the answer as well. Please correct me if I am missing something. 
 [/INST] The functions such that $f(x)=f(2-x)$ for all $x$ are the functions with a symmetry around the vertical line $x=1$ (we can see this solving $x=2-x$, giving us $x=1$).
Therefore we can exclude $B)$ and $C)$ since the graph is a line and they can't have this kind of symmetry (unless they are horizontal lines, which is not the case).
$A)$ has as symmetry axis the line $x=-1$
The symmetry axis for $D$ is indeed $x=1$ and if you calculate $f(2-x)=3(2-x)(2-x-2)=3(2-x)(-x)=3x(x-2)=f(x)$
For $E)$ $f(2-x)=(2-x)^2(2-(2-x))^2=(2-x)^2x^2=f(x)$
So the correct answers are $D)$ and $E)$
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