# How do you check which intervals a cubic function will increase and in which intervals it will decrease?

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.

• HINT: Differentiate it - that gives you the slope at any point along the function Commented Mar 15, 2016 at 14:19
• @Mufasa I already differentiated it, that's how i got that the differential is zero at {-2,3}. What i don't know is which of the three intervals i get would the function be strictly increasing or strictly decreasing in. Commented Mar 15, 2016 at 14:21
• The derivative is $12x^2-12x-72=12(x^2-x-6)=12(x-3)(x+2)$ Now use this to determine where the slope is negative, zero and positive. Negative slope implies function is decreasing, positive slope implies the function is increasing. Commented Mar 15, 2016 at 14:23
• Hint: in order for the function to switch from increasing to decreasing, we have to pass a point with derivative zero Commented Mar 15, 2016 at 14:25

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.

• Can you just explain it in simple terms. Like compute the derivative and set it equal to 0, and then solve (or whatever one would do to solve this). Commented Jun 28, 2023 at 18:21
• That's essentially what this answer does: Since we don't care about the roots, just whether they are real or not, we don't need to bother computing the roots---it's enough just to compute the discriminant, $\Delta$. Commented Jun 28, 2023 at 19:48
• Invoking the discriminant is more confusing (at least for me) since I don't know what that is. Commented Jun 28, 2023 at 19:58
• The discriminant is the quantity under the radical quadratic formula, so something you'd have to compute anyway to find the roots. Commented Jun 28, 2023 at 20:39

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".