What about ?
Or even ?
No pattern.
I would like to verify if cubic functions always have the following pattern
.
And also, does it always have a positive-negative-positive-negative or negative-positive-negative-positive pattern.
These facts are essential to help me solve inequalities with cubic polynomials. Eg. x(x+1)(x-1)<0.
As long as I establish the sign before 1 root. I can deduce the pattern, hence saving much of my time.
I need at least 3 experts to verify.
For a cubic where , we can make some observations (provable)
1) there is at least one real root
2) there are at most three real roots
3) if a > 0, then and
4) if a < 0, then and
You may not understand (3) and (4).... anyway for the example you gave
x(x+1)(x-1)<0
if you look at the equality (instead of inequality)
x(x+1)(x-1)=0
you should be able to see immediately there are three real roots, 0, -1 and 1. Thus we pay attention to four regions of the real number line, which are
, , ,
Now you can test out values in each interval, or you can use (3) above and a little logic to deduce that the satisfying intervals are and .
Note that if a cubic has three real roots then the function must change signs on either side of each root (locally); if it has two real roots then one must be a double root (the function does not change sign there); if one real root then the function changes sign there.
Example of two real roots: f(x) = (x-2)(x-3)^2 (graph it to get a visual)
Example of one real root: f(x) = (x-5)^3 (graph it) -- this one has a triple root
Another example of one real root: f(x) = (x-2)(x-3)(x-4) + 20 (this does not have a triple root)
That depends very much on how strictly we define "looks like."
Some possible counter-examples:
f(x)=x^3 (it only changes direction once)
f(x)=-(x-1)(x-2)(x-3) (it is upside down)
f(x)=10x(x-1)(x+1) (if you use the same scale for x- and y-axes, it will be a very "sharp" squiggle)