The line through the two local extremum points of a cubic

Content updated: October 22, 2026

The line through a cubic's two local extremum points can be found from their coordinates or from the remainder on division by the derivative. The remainder method has this geometric meaning only after verifying that both extrema exist.

On this page: Idea · Method · Examples · Practice · Further learning

1. Why the remainder passes through both points

Let $f(x)=ax^3+bx^2+cx+d$, $a\ne0$. The function has two local extrema exactly when $D=b^2-3ac>0$. Let $x_1,x_2$ be the distinct roots of $f^{\prime}$. Polynomial division has the form $f=qf^{\prime}+R$, where $R$ has degree at most $1$. Substitution of $x_i$ makes the derivative term $0$, so $f(x_i)=R(x_i)$. Both points $(x_i;f(x_i))$ lie on $y=R(x)$.

Since $x_1\ne x_2$, the two points are distinct and determine a unique line; thus $y=R(x)$ is the required line. This is a polynomial division identity, not division of the value $f(x)/f^{\prime}(x)$ where the derivative is $0$.

2. General formula and checking it

Dividing once for a general cubic gives

Thus, if $D>0$, the required equation is

Check by multiplying the quotient by the derivative and adding the remainder: all four coefficients of $f$ must be recovered. If $D\le0$, the remainder formula is still an algebraic expression, but there are no two distinct real local extremum points giving it this geometric interpretation. Its slope is $-2D/(9a)$ and therefore differs from $0$ when both extrema exist.

3. Two examples, each with two methods

Example 1. A cubic without a parameter

Let $f=x^3-3x+2$. Its derivative $3x^2-3$ has simple roots $1,-1$. Coordinate method: $f(1)=0$ and $f(-1)=4$, so the points are $(1;0)$ and $(-1;4)$. The slope is $(4-0)/(-1-1)=-2$, giving $y=-2x+2$, or $2x+y-2=0$.

Remainder method: check $f=(x/3)f^{\prime}+(-2x+2)$ by multiplying out. At both derivative roots, $f=-2x+2$, giving the same line. The horizontal coordinates differ, so the slope denominator is not $0$.

Example 2. A parameter and distinctness

Let $f=2x^3+3mx^2-12m^2x+m+2$. Its derivative $6x^2+6mx-12m^2=6(x-m)(x+2m)$ has two simple roots exactly when $m\ne0$. The reduced discriminant $81m^2$ confirms the same condition.

Coordinate method: the two local extremum points are $(m;-7m^3+m+2)$ and $(-2m;20m^3+m+2)$. For $m\ne0$, the slope is $27m^3/(-3m)=-9m^2$. Substitution of a point gives constant term $2m^3+m+2$. Hence

Remainder method: the correct division identity is

Multiply back: the quotient–derivative product has coefficient of $x$ equal to $-3m^2$; adding remainder term $-9m^2x$ recovers $-12m^2x$ in the problem. Thus the slope must be $-9m^2$. For $m=0$, the function becomes $2x^3+2$, with no two local extrema; the remainder line cannot be accepted as their joining line.

4. Practice exercises

1. Find the line through the two local extremum points of $x^3-3x$.
2. Find the line through the two local extremum points of $2x^3-6x+1$.
3. Can the remainder line for $x^3+x$ be called the line through two local extremum points?

5. Common mistakes

• Omitting the distinct-derivative-root condition.
• Copying a remainder without checking by multiplication.
• Dividing values by the derivative where it equals $0$.
• Keeping a geometric answer at a parameter where the two points coincide.

6. Related lessons

• Conditions for local extrema (English lesson)
• Coordinate conditions on local extrema (English lesson)
• Local extrema (English lesson)

Original exercise material on Scribd.