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?
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?
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.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 confusedThis 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.
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.