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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.

user289143
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ANANYA AKHOURI
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1 Answers1

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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 $E)$

user289143
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  • Thank you for the answer @user289143 and thank you so much for suggesting the edit as well! Much appreciated! It reads much easier and better now! I'm afraid I have a few more doubts. I didn't understand why we're putting x=2-x. It is a function in x, so it can be anything, isn't it? Can't we use x^2 = (2-x)^2 also then? Or is it a way to find out symmetry? Sorry I am not aware if it's a method. Also, how did we find that the symmetry axis for D is x=1? – ANANYA AKHOURI Jun 29 '19 at 11:08
  • @ANANYAAKHOURI I put $x=2-x$ to find the $x$ that belongs to the axis of symmetry. In general when a function $f$ is symmetric around a vertical line $x=a$ it means that $f(a-x)=f(a+x)$ and substituting $a-x$ instead of $x$ we get $f(x)=f(2a-x)$ so we can see in our case that $a=1$. Regarding $D)$: this function is a parabola $f(x)=ax^2+bx+c$ and the axis of symmetry is $x=-b/2a$ – user289143 Jun 29 '19 at 11:28
  • Thank you for the explanation! I understood the parabola part and that if a function is symmetric around $x=a$, it means that $f(a−x)=f(a+x)$. However, I did not understand one thing. If we substitute $a-x$ instead of $x$, won't we get $f(a-x)=f(2-a+x)$ ? You have written that we will get $f(x) = f(2a-x)$. I did not understand how. I guess I need to study the symmetry of functions properly. – ANANYA AKHOURI Jun 30 '19 at 07:32
  • You have $f(a-x)=f(a+x)$. Now let's put $a-x$ instead of $x$: $f(a-(a-x))=f(a+(a-x))$ i.e. $f(x)=f(2a-x)$ – user289143 Jun 30 '19 at 07:55
  • I got it! Thanks a lot for the clarification! – ANANYA AKHOURI Jul 01 '19 at 08:37