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?
andyrk said: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?
How did you reduce it down to that? Can you show me? I am unable to get to that point.Shyan said: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.
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.andyrk said: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?
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 confusedPeroK said: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.
andyrk said: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.