How Do You Check Which Intervals a Cubic Function Will Increase and in Which Intervals It Will Decrease?
$\Begingroup$ I Was Trying to Find the Intervals in Which the Cubic Function $4X^3 -6X^2 -72X + 30$ Would Be Strictly Increasing and Strictly Decreasing. I...
I was trying to find the intervals in which the cubic function $4x^3 -6x^2 -72x + 30$ would be strictly increasing and strictly decreasing.
I managed to get the fact that at the values {-2,3} the differential of the function is zero. However this divides the function in three intervals, how can i know which intervals the function increases in and which intervals the function decreases in?
Note: I know you could simply plot the function. I was hoping for a more analytical method.
2 Answers
Whether a differentiable function is increasing or decreasing (or stationary) at a point is (essentially) determined by the sign of its derivative. For a cubic polynomial function $$p(x) = a x^3 + b x^2 + c x + d ,$$ the derivative is $$p'(x) = 3 a x^2 + 2 b x + c .$$ Thus, the character of the roots of $p'$, that is, the critical points of $p$, is determined completely by the discriminant $\Delta := ((2 b)^2) - 4 (3a) (c) = 4 b^2 - 12 a c$ of $p'$:
- If $\Delta > 0$, then $p'$ has two distinct roots, $r_- < r_+$. If $a > 0$ (in particular, if it is monic), then since $\lim_{x \to \pm \infty} p'(x) = +\infty$, we conclude that $p$ is increasing where $x < r_-$ and $x > r_+$, decreasing where $r_- < x < r_+$, and hence has a local maximum at $r_-$ and a local minimum at $r_+$.
- If $\Delta = 0$, then $p'$ has a double root, $r$. If $a > 0$, we conclude that $p$ is increasing where $x \neq r$ and that $p'(r) = 0$, so that $p$ has an inflection point at $r$, and has no minimum or maximum. (In fact, $p$ is strictly increasing everywhere, as $x < y$ implies $p(x) < p(y)$.)
- If $\Delta < 0$, then $p'$ has no roots. If $a > 0$, $p$ is increasing everywhere.
(All of these statements are reversed appropriately when $a < 0$.)
Example For our polynomial $p(x) = 4 x^3 - 6 x^2 - 72 x + 30,$ computing gives $\Delta = 3600 > 0$, and we are in the first case.
You could look at the leading coefficient, or you can compute the derivative at a point in one of your intervals (or all, depending on what you want to assume).
Let's look at this in a little more depth. We have the function
$$f(x)=4x^3-6x^2-72x+30$$
with derivative
$$f'(x)=12x^2-12x-72$$
We know the function is either going "up-down-up" or "down-up-down" (this you can say if you found two distict zeroes in the derivative- if not, then the cubic goes only "up" or only "down"), since you've found $f'(-2)=f'(3)=0$. We can look at the leading coefficient to check where the function goes up and down (if the leading coefficient is positive, then it is "up-down-up" and otherwise, it is "down-up-down". I might make this a little more intuitive if I say: in, for example, $x^3-x^2$, the $x^3$ will always win of $-x^2$, so in the end, the function will increase). We can also compute the derivative in a specific point in one of the intervals $(-\infty,-2)$, $(-2,3)$ or $(3,\infty)$, for example, in $(-2,3)$, to get
$$f'(0)=-72$$
to know that the function is decreasing on that interval, and so it must be "up-down-up".
Sarah Jenkins is a veteran tech journalist with over 12 years of experience covering artificial intelligence, mobile innovations, and digital ethics. Her insights have appeared in leading technology publications worldwide.
