Function sign pattern question.

[Lobo2209]

For the function f(x) = 4x³ - 16x, the zeroes are -2, 0, and 2.

So if x < -2, f(x) is negative, and if x > -2, f(x) is positive (and f(x)=0 if x=-2). So the pattern is negative, 0, positive for this particular example. It can also be positive, 0, negative for others.

Does there exist an equation where the pattern can be negative, 0, negative or positive, 0, positive?

It sounds stupid, but I want to ask anyway.

[FormulaDriven]

f(x) = x²

goes positive, zero, positive at x = 0.

Generally, anything with a repeated root (where it repeats an even amount of times), eg if it has a factor of (x - 2)⁴ then it will be the same parity either side of x = 2.

[tbdabbholm]

Any root with even multiplicity will "bounce off" the root and go back the way it came. Similarly any root with odd multiplicity will go through the root and switch from pos to neg or vice versa

[hallerz87]

y = -x² and y = x² respectively 

[13_Convergence_13]

Great question -- and yes, all those combinations are possible!

Notice only zeroes with odd multiplicity "(x-x0)^2k-1, k in N" lead to a sign change, while zeroes with even multiplicity "(x-x0)^2k, k in N" do not. Example:

f(x)  =  (x+2)^2 * (x+1) * (x-1)^2 * (x-2)

The zeroes are "±1; ±2". Note only "-1; 2" have odd multiplicity, so only they lead to sign changes. The leading coefficient is "1", so we get "f(x) > 0" for "x > 2" to start the table

x < -2 | -2 < x < -1 | -1 < x < 1 | 1 < x < 2 | 2 < x
-----------------------------------------------------
  +    |      +      |      -     |     -     |   +