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F(x) symmetric about the line x=2

  1. Feb 2, 2015 #1
    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?
     
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  3. Feb 2, 2015 #2

    ShayanJ

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    Change the variable to y=x-2 and see whether the resulting function is even or odd or none!
     
  4. Feb 2, 2015 #3
    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.
     
  5. Feb 2, 2015 #4

    ShayanJ

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    ## 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).
     
  6. Feb 2, 2015 #5
    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?
     
  7. Feb 2, 2015 #6

    ShayanJ

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    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.
     
  8. Feb 2, 2015 #7
    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?
     
  9. Feb 2, 2015 #8

    ShayanJ

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    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.
     
  10. Feb 2, 2015 #9

    PeroK

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    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.
     
  11. Feb 2, 2015 #10
    How did you reduce it down to that? Can you show me? I am unable to get to that point.
     
  12. Feb 2, 2015 #11

    ShayanJ

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    ## 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##
     
  13. Feb 2, 2015 #12
    And what is the reason that we transformed x to x+2?
     
  14. Feb 2, 2015 #13

    ShayanJ

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    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.
     
  15. Feb 2, 2015 #14
    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?
     
  16. Feb 2, 2015 #15

    ShayanJ

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  17. Feb 2, 2015 #16
    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.
     
  18. Feb 2, 2015 #17

    PeroK

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    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.
     
  19. Feb 2, 2015 #18
    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
     
  20. Feb 2, 2015 #19

    ShayanJ

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    ## z=x-2 \Rightarrow x=z+2 ## so ## f(x)=f(z+2) ##!(Forget the confusing ## x\rightarrow x+2 ##!)
     
  21. Feb 2, 2015 #20
    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.
     
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