Parameters for monotonicity on a specified interval
Determine parameter values using derivative signs while checking degree reductions, open-interval boundaries, and excluded poles.
Content updated: October 22, 2026
Finding parameter values for monotonicity on an interval requires two checks: the function is defined throughout the interval, and its derivative has the required sign at every point. Values causing a degree reduction require separate treatment.
Objectives: handle cubic and linear fractional functions, retain boundary cases, and check the location of undefined points.
Prerequisites: monotone functions (English lesson); signs of quadratic polynomials and ordering their roots.
On this page: Idea · Method · Examples · Practice · Further learning
1. Check both the domain and the sign
This lesson uses “increasing” and “decreasing” strictly. For the nonconstant polynomials and rational functions below, a nonnegative or nonpositive derivative on an interval, together with the absence of constant subintervals, corresponds to strict monotonicity. A negative derivative even at one point disproves increase for a differentiable function; similarly, a positive derivative disproves decrease.
Before dividing by a coefficient containing $m$ or applying a discriminant criterion, examine values making that coefficient $0$. For a rational expression, a domain hole inside the required interval means the function is not defined on the whole interval, even if a factor could be canceled algebraically.
2. Two derivative models
For $f(x)=ax^3+bx^2+cx+d$, $a\ne0$, set $D=b^2-3ac$, the reduced discriminant of $f^{\prime}=3ax^2+2bx+c$. If $D<0$, the derivative always has the sign of $a$. If $D=0$, it has the sign of $a$ except at a double zero and does not change sign. If $D>0$, order the roots as $r_1<r_2$: the derivative has the sign of $a$ outside $(r_1;r_2)$ and the opposite sign inside.
Thus a cubic increases on $\mathbb R$ exactly when $a>0,D\le0$, and decreases on $\mathbb R$ exactly when $a<0,D\le0$. On a smaller interval, $D\le0$ is unnecessary: the required interval may lie wholly in one suitable sign region. A derivative root at the endpoint of an open interval is allowed.
For $F(x)=(ax+b)/(cx+d)$, suppose the denominator is not identically $0$. Its derivative is
If $ad-bc\ne0$, the sign depends only on the numerator. For $c\ne0$, exclude the pole $-d/c$ and consider each interval separately. For $c=0$, $d\ne0$, the function is affine on $\mathbb R$, with slope $a/d$. If $ad-bc=0$, it is constant on its domain intervals and is not strictly monotone.
3. Seven worked examples
Example 1. Increase on the whole of R
Let $f=x^3+(m+1)x^2+(m^2-1)x+9$. Then $f^{\prime}=3x^2+2(m+1)x+m^2-1$ and $D=-2m^2+2m+4=-2(m+1)(m-2)$. The leading coefficient is positive, so $D\le0$ is equivalent to $m\le-1$ or $m\ge2$. Both boundary values are included: the derivative has a double zero, but no constant interval occurs.
Video accompanying the original lesson (Vietnamese-language video).
Example 2. Decrease and a degree reduction
Let $f=(m^2-1)x^3/3+(m+1)x^2-x$. For $m\ne\pm1$, its derivative is quadratic with leading coefficient $m^2-1$ and reduced discriminant $2m^2+2m$. Nonpositivity on $\mathbb R$ is equivalent to $|m|<1$ and $m(m+1)\le0$, giving $-1<m\le0$.
Treat $m=-1$ separately: the function is $-x$, which decreases. For $m=1$, it is $2x^2-x$, which does not decrease on all of $\mathbb R$. Combining cases gives $-1\le m\le0$.
Video accompanying the original lesson (Vietnamese-language video).
Example 3. Increase on a half-line
Consider the function
Require increase on $(2;+\infty)$. We have
If $m\le1$, the roots satisfy $2m\le m+1\le2$, so the derivative is positive for every $x>2$. If $m>1$, their order is $2<m+1<2m$; the interval between them lies in the required interval, and the derivative is negative there. Thus the exact condition is $m\le1$, including the double-zero case $m=1$.
Video accompanying the original lesson (Vietnamese-language video).
Example 4. Decrease on a finite interval
For the function in Example 1, require decrease on $(-1;0)$. Set $q=f^{\prime}$, a quadratic with leading coefficient $3>0$. The condition $q\le0$ on the open interval is equivalent to $q(-1)\le0$ and $q(0)\le0$: necessity follows from continuity, and sufficiency because the parabola lies below the chord joining the endpoint values on this closed interval. It lies strictly below that chord in the interior, so the derivative is negative there.
Intersecting $0\le m\le2$ and $-1\le m\le1$ gives $0\le m\le1$. Boundary cases with derivative zeros at endpoints remain valid.
Video accompanying the original lesson (Vietnamese-language video).
Example 5. Checking a quadratic vertex on a half-line
Let $f=x^3-(m+1)x^2+(m^2-1)x+9$ and require increase on $(1;+\infty)$. The derivative $q=3x^2-2(m+1)x+m^2-1$ has vertex at $v=(m+1)/3$.
If $m\le2$, the vertex lies at or to the left of $1$; the minimum on $[1;+\infty)$ is $q(1)=m(m-2)$. Nonnegativity gives $m\le0$ or $m=2$. If $m>2$, the vertex is inside the interval and $D=-2(m+1)(m-2)<0$, so $q>0$ on all of $\mathbb R$. Hence the answer is $m\le0$ or $m\ge2$.
Video accompanying the original lesson (Vietnamese-language video).
Example 6. The length of a decreasing interval
Let $f=x^3+mx^2+(m^2-6)x+m+1$. Its derivative $q=3x^2+2mx+m^2-6$ has two distinct roots when $D=18-2m^2>0$, or $|m|<3$. Call them $r_1<r_2$; the maximal interval of decrease is $(r_1;r_2)$, whose length is
If this maximal decreasing interval must have length exactly $2$, solve $18-2m^2=9$ to obtain $m=\pm3/\sqrt2$. If the requirement is only that some subinterval of length $2$ supports decrease, the maximal length must be at least $2$, giving $|m|\le3/\sqrt2$. Distinguish these two meanings of “an interval of length 2.”
Video accompanying the original lesson (Vietnamese-language video).
Example 7. A rational function and the position of its pole
Let $F(x)=(mx+4)/(x+m)$, whose domain excludes $-m$ and whose derivative is $(m^2-4)/(x+m)^2$. Increase on each domain interval requires $m^2-4>0$, or $m<-2$ or $m>2$. For $m=\pm2$, the function is constant on its domain, so these values are excluded.
For decrease on $(-1;0)$, first require $-2<m<2$. Also, the pole $-m$ must not belong to that open interval: $-m\le-1$ or $-m\ge0$, equivalent to $m\ge1$ or $m\le0$. Intersecting the conditions gives
The boundary cases $m=0$ and $m=1$ are included because the poles then lie at the endpoints $0$ and $-1$, outside the required open interval. Monotonicity on each interval is not monotonicity on their whole union across the pole.
Video accompanying the original lesson (Vietnamese-language video).
Original video on the linear fractional model (Vietnamese-language video).
4. Practice exercises
- Find $m$ such that $x^3-3mx$ increases on $\mathbb R$.
- Find $m$ such that $(mx+1)/(x+2)$ increases on $(0;+\infty)$.
- Find $m$ such that $x^3-3x$ decreases on $(m;m+1)$.
Show answers and hints
- The derivative $3(x^2-m)$ is nonnegative for every $x$ exactly when $m\le0$; the function has no constant interval and is therefore strictly increasing.
- The pole $-2$ is outside the required interval; the derivative numerator is $2m-1$, so $m>1/2$.
- The maximal decreasing interval is $(-1;1)$. Require $m\ge-1$ and $m+1\le1$, hence $-1\le m\le0$.
5. Common mistakes
- Omitting degree-reduction cases before applying a discriminant criterion.
- Replacing the pole $-m$ incorrectly by $m$.
- Excluding derivative roots at the boundary of an open interval.
- Confusing a maximal monotonicity interval with one of its subintervals.
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