Finding value of $f(4)$ for a cubic polynomial under given derivative and sign conditions

Question

Let $f(x)$ be a monic cubic polynomial (i.e., leading coefficient is $1$) such that it satisfies the following conditions:

1. There exists no integer $k$ such that $f(k+1) \cdot f(k-1) < 0$

2. $f^\prime\left(-\frac{1}{4}\right) = -\frac{1}{4}$

3. $f^\prime\left(\frac{1}{4}\right) > 0$

We are asked to find the value of $f(4)$.

Options:

• $(a) \quad 81.50$
• $(b) \quad 82$
• $(c) \quad 81$
• $(d) \quad 82.50$

What we’ve tried so far:

• Since $f(x)$ is a monic cubic, we assumed it to be of the form $f(x) = x^3 + ax^2 + bx + c$ and hence, $f^\prime(x) = 3x^2 + 2ax + b$
• We used the derivative conditions at $x = -\frac{1}{4}$ and $x = \frac{1}{4}$ to form equations and inequalities involving $a$ and $b$.
• We also considered condition (1), which restricts the sign change behavior of the polynomial at integer points, helping us infer the nature of root placements and sign pattern of $f(x)$.

We believe the key lies in reconstructing the polynomial using these conditions to find $f(4)$, but we’d love some help verifying the logic and final computation.

Any help or insight is appreciated!

Answer 1

$\DeclareMathOperator{\sgn}{sgn}$ Let $f(x)=x^3+ax^2+bx+c$. Condition $f'(-\frac{1}{4})=-\frac{1}{4}$ implies $a=2b+\frac{7}{8}$ thus $f(x)=x^3+(2b+\frac{7}{8})x^2+bx+c$. Let $F:=\{\sgn f(k)\}_{k\in\mathbb{Z}}$ , $A:=\{F_{2k}\}_{k\in\mathbb{Z}}$ , $B:=\{F_{2k+1}\}_{k\in\mathbb{Z}}$.
$\not\exists k\in\mathbb{Z},f(k-1)f(k+1)<0\iff\forall k\in\mathbb{Z}, X_kX_{k+1}\ge0$ where $X\in\{A,B\}$.
We know that: $\lim_{x\to-\infty}f(x)=-\infty$ , $\lim_{x\to\infty}f(x)=\infty$ , so $\exists k_1,k_2\in\mathbb{Z},X_{k_1}X_{k_2}<0$.
It's not hard to deduce that: $$(\forall k\in\mathbb{Z}, X_kX_{k+1}\ge0)\wedge(\exists k_1,k_2\in\mathbb{Z},X_{k_1}X_{k_2}<0)\implies(\exists k\in\mathbb{Z},X_k=0)$$ Then $f(x)$ must have at least 2 integer roots but $f(x)$ cannot have 3 integer roots (distinct or repeated) since that would imply $(2b+\frac{7}{8}\in\mathbb{Z})\wedge(b\in\mathbb{Z})$ ,which is false, so $f(x)$ has exactly 2 non-repeated integer roots $m<n$ and 1 non-integer root $x_0$, this implies $\exists !k\in\mathbb{Z},X_k=0$.
We have:$$(\forall k\in\mathbb{Z}, X_kX_{k+1}\ge0)\wedge(\exists !k\in\mathbb{Z},X_k=0)\implies X=\{...,-1,-1,0,1,1,...\}$$ Then we must have $n=m+1$ or $n=m+3$ because if $n$ and $m$ were too separated, the sequence $F$ would be something like: $F=\{...,-1,-1,0,-1,1,-1,1,0,1,1,...\}$ and there would be too many roots. Now there are 2 cases: case $n=m+3$ and case $n=m+1$.

Case 1: $n=m+3$
$F=\{...,-1,-1,0,-1,1,0,1,1,...\}$, $x_0\in(m+1,m+2)$.
Use Vieta's formula: $$\begin{cases}m+n+x_0=-2b-\frac{7}{8}\\ mn+mx_0+nx_0=b\end{cases}\implies x_0=-\frac{2m^2+8m+\frac{31}{8}}{4m+7}$$ Solve the following inequality for integers: $$m+1<-\frac{2m^2+8m+\frac{31}{8}}{4m+7}<m+2$$ we find out $m=-1,n=2,x_0=\frac{17}{24}$ , so $f(x)=(x+1)(x-2)(x-\frac{17}{24})$, this function does not satisfy $f'(\frac{1}{4})>0$ (and actually $f(-2)f(0)<0$).
->No solution in case 1, case closed.

(Update: I just realized you can skip the whole calculation in case 1. Notice that $F_{m-1}F_{m+1}>0$ and $F_m=0$, this contradicts the fact that m is a single root)

Case 2: $n=m+1$
$F=\{...,-1,-1,0,0,1,1,...\}$
We must have $f(\lfloor x_0\rfloor)f(\lfloor x_0\rfloor+1)\not>0$ otherwise there will be 2 distinct roots or 1 repeated root in $(\lfloor x_0\rfloor,\lfloor x_0\rfloor+1)$ thus $m-1<x_0<m+2$.
Again, by Vieta's formula: $x_0=-\frac{2m^2+4m+\frac{15}{8}}{4m+3}=-\frac{1}{2}m-\frac{5}{8}$

Solving $m-1<-\frac{1}{2}m-\frac{5}{8}<m+2$, we find out $m=-1$ or $m=0$.
Therefore: $$(m,n,x_0)\in\{(-1,0,-\frac{1}{8}),(0,1,-\frac{5}{8})\}$$ $$\implies f(x)=(x+1)x(x+\frac{1}{8})\ \text{or}\ f(x)=x(x-1)(x+\frac{5}{8})$$ and $f(x)=(x+1)x\left(x+\frac{1}{8}\right)$ satisfy all conditions thus it is the only solution.

Answer: $f(4)=82.5$

Answer 2

We also considered condition (1), which restricts the sign change behavior of the polynomial at integer points, helping us infer the nature of root placements and sign pattern of $f(x)$.

This statement is very helpful in determining the roots and shape of $f(x)$. Specifically, condition (1) implies there must be at least one even and one odd integer root, let's call them $O$ and $E$, along with another root, which we'll call $r$. If $|O-E|>1$, then at least one of $f(E-1)\cdot f(E+1)$ or $f(O-1)\cdot f(O+1)$ will be negative, which is a contradiction. Thus we can rewrite $f(x)$ as $f(x)=(x-r)(x-s)(x-s-1)$ for some integer $s$. Note that because of the conditions mentioned above, $r$ must be in the interval $[s-1,s+2]$

Using condition (2), we can use $f'(-\frac{1}{4})=-\frac{1}{4}$ to solve for $r$ in terms of $s$ to get $$r=-\frac{8s^2+16s+7}{16s+12}~.$$ To satisfy the condition mentioned above, we need $$-\frac{8s^2+16s+7}{16s+12}-s=-\frac{24s^2+28s+7}{16s+12}\in[-1,2]~,$$ which is somewhat easily solved by noticing that this expression has slant asymptote $y=-\frac{3}{2}x-\frac{5}{8}$ and vertical asymptote $x=-\frac{3}{4}$ to give that only $s=-1$ and $s=0$ satisfy the conditions.

Finally, using condition (3), we can eliminate $s=0$, which gives $r=-\frac{1}{8}$ and thus $f(x)=x(x+1)(x+\frac{1}{8})$. Plugging in $4$, we get $f(4)=\boxed{\textbf{(d) }82.50}$