F(x) symmetric about the line x=2

1. Feb 2, 2015

andyrk

Why is the function: y = f(x) = a(x-1)(x-2)(x-3) symmetrical about the line x = 2? I mean how can we be sure that it is? Is there any method to check it?

2. Feb 2, 2015

ShayanJ

Change the variable to y=x-2 and see whether the resulting function is even or odd or none!

3. Feb 2, 2015

andyrk

You mean x = x-2 right? It doesn't stay the same on doing that. So it should not be symmetrical about x = 2, but it is. And I am not able to see how that happens.

4. Feb 2, 2015

ShayanJ

$f(x)=a(x-1)(x-2)(x-3) \rightarrow f(y)=a(y+1)y(y-1)$
$f(-y)=a(-y+1)(-y)(-y-1)=-a(y-1)y(y+1)=-f(y)$
So f(y) is odd which means f(x) is not symmetric around x=2 but is not that much asymmetric because we have f(2-x)=-f(x-2).

5. Feb 2, 2015

andyrk

Oh no. I posted the wrong function. The function is: y = f(x) = a(x4/4 -2x3 + 11x2/2 - 6x) + 1

And this function is said to be symmetrical about the line x = 2. But I am unable to see how?

6. Feb 2, 2015

ShayanJ

Its the same trick. Just do the tranformation $x\rightarrow x+2$ and check whether the resulting function is even. If its even, then the original function is symmetric around x=2.

7. Feb 2, 2015

andyrk

They are not the same, i.e. after changing x to x+2 in f(x), f(x) ≠ f(-x). But what I am reading, it says that f(x) is symmetrical about x=2 and I am still wondering that how would I go about proving it?

8. Feb 2, 2015

ShayanJ

You're doing something wrong. You should be able to reduce f(x+2) to $\frac 1 2 a[x^2(\frac 1 2 x^2-1)-4]+1$ which is even.

9. Feb 2, 2015

PeroK

Another way to look at it is as follows:

Imagine starting at x = 2 and moving the same distance, d, to the right and left (d > 0). So, to the right we have 2 + d and to the left we have 2 - d.

Now, if f is symmetrical about x = 2, then f(2-d) = f(2+d) for all d. You could try that approach.

10. Feb 2, 2015

andyrk

How did you reduce it down to that? Can you show me? I am unable to get to that point.

11. Feb 2, 2015

ShayanJ

$f(x+2)=a[ \frac 1 4 (x+2)^4-2(x+2)^3+\frac{11}{2} (x+2)^2-6(x+2)]+1=\\ a(x+2)[ \frac 1 4 (x+2)^3-2(x+2)^2+\frac{11}{2} (x+2)-6]+1=\\ a(x+2)(\frac 1 4 x^3+\frac 3 2 x^2+3x+2-2x^2-8x-8+\frac{11}{2}x+5)+1=\\ \frac 1 2 a(x+2)(\frac 1 2 x^3-x^2+x-2)+1=\frac 1 2 a (\frac 1 2 x^4+x^3-x^3-2x^2+x^2+2x-2x-4)+1=\\ \frac 1 2 a (\frac 1 2 x^4-x^2-4)+1$

12. Feb 2, 2015

andyrk

And what is the reason that we transformed x to x+2?

13. Feb 2, 2015

ShayanJ

Simple. When we say a function is even, we mean its symmetric around x=0. So if a function is symmetric around $x=a \neq 0$, it means if we move the origin to x=a, the resulting function would be even.

14. Feb 2, 2015

andyrk

Right. So if we move the coordinates of origin from (0,0) to (2,0) shouldn't the abscissa of a point in the new axes, change from x to x-2 and not x+2?

15. Feb 2, 2015

ShayanJ

16. Feb 2, 2015

andyrk

But we aren't shifting the curve as in your link but instead we are shifting the coordinate axes. The curve stays where it was. The axes are what shift.

17. Feb 2, 2015

PeroK

This is one reason why it's much better to change the variable name - until, like Shyan, you've mastered this. Let's use z.

If we have z = x + 2, then x = 2 maps to z = 4, which is not what we want.

But, if we have z = x - 2, then x = 2 maps to z = 0, which is what we want. The origin of the z-variable is at x = 2.

In fact, if I'm honest, I always prefer to change the name of the variable to avoid the mistake you just made.

18. Feb 2, 2015

andyrk

Wait. So you are saying that f(x) should change to a new variable so that it becomes f(z), right? But that would again mean f(x-2) and not f(x+2).. got me confused

19. Feb 2, 2015

ShayanJ

$z=x-2 \Rightarrow x=z+2$ so $f(x)=f(z+2)$!(Forget the confusing $x\rightarrow x+2$!)

20. Feb 2, 2015

andyrk

It just simply fails to make sense to me because of the simple logic that in older coordinate plane, suppose x (any number in domain of f(x)) mapped to f(x) = y (say). Now, in the new coordinate system, f(xold) = f(xnew) provided that xnew = xold - 2. So shouldn't we evaluate f(x-2) instead of f(x+2)?

For example, earlier xold = 2 gave f(2) = c. Now xnew = 0 (i.e. xold-2, where xold = 2 ) would give the result c.