Find range of values of $a,b,c$ such that $ax^2+bx+c$ satisfies given conditions

Question

Let $y = f(x) = ax^2+bx+c$ where $a\neq0$. Find the range of values $a,b$ and $c$ such that it satisfies the following:

• $0 \le f(x) \le 1 \quad \forall \space x \in [0,1]$

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What I have found so far?

1. For $x=0$, we get $0 \le c \le 1$

2. For $x=1$, we get $0 \le a+b+c \le 1$

3. If roots exist, then they should lie outside the interval $[0,1]$.

   Therefore, ${{-b - \sqrt{D}}\over{2a}} < 0$ and ${{-b + \sqrt{D}}\over{2a}} >1$. I tried to solve these but didn't get any fruitful results.

   EDIT 1: As explained in comments by @insipidintegrator, this is only for distinct roots.

4. Vertex of quadratic $f(x)$ is $\big( \frac{-b}{2a}, -\frac{D}{4a}\big)$. If the x-coordinate of the vertex lies between 0 and 1, then y-coordinate should also be between 0 and 1. But I have no idea how to proceed from here.

I want to find some lower/upper limits for $a,b,c$ or any kind of relation between them. Any solution or hints for solving this problem is greatly appreciated.

Answer 1

We may obtain the necessary and sufficient conditions for $$a, b, c \in \mathbb{R}, ~ 0 \le ax^2 + bx + c \le 1, \forall x \in [0, 1].$$

Fact 1: Let $A, B, C \in \mathbb{R}$. Then $$Ax^2 + Bx + C \ge 0, ~ x \ge 0 \iff A \ge 0, ~ C \ge 0, ~ B \ge -\sqrt{4AC}.$$ (The proof is given at the end.)

Fact 2: Let $A, B, C \in \mathbb{R}$. Then $$Ax^2 + Bx + C \ge 0, ~ x\in [0, 1] \iff \left\{\begin{array}{l} A + B + C \ge 0\\[5pt] C \ge 0\\[5pt] B + 2C \ge - \sqrt{4(A + B + C)C}. \end{array} \right. $$ (The proof is given at the end.)

Now, using Fact 2, we have $$ax^2 + bx + c \ge 0, ~ x \in [0, 1] \iff \left\{\begin{array}{l} a + b + c \ge 0\\[5pt] c \ge 0\\[5pt] b + 2c \ge - \sqrt{4(a + b + c)c} \end{array} \right.$$ and $$ - ax^2 - bx - c + 1 \ge 0,~ x \in [0, 1]$$ $$\iff \left\{\begin{array}{l} -a - b - c + 1 \ge 0\\[5pt] - c + 1 \ge 0\\[5pt] - b + 2(-c + 1) \ge - \sqrt{4(-a - b - c + 1)(-c + 1)}. \end{array} \right.$$

Thus, we have $$0 \le ax^2 + bx + c \le 1, ~ x\in [0, 1]$$ $$\iff \left\{\begin{array}{l} 0 \le a + b + c \le 1\\[5pt] 0 \le c \le 1\\[5pt] b + 2c \ge - \sqrt{4(a + b + c)c}\\[5pt] b + 2c \le 2 + \sqrt{4(-a - b - c + 1)(-c + 1)}. \end{array} \right.$$

$\phantom{2}$

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Proof of Fact 1:

“$\Longleftarrow$”: $\quad$ For $x\ge 0$, we have $Ax^2+Bx+C\ge Ax^2 - \sqrt{4AC}\ x + C = (x\sqrt{A} - \sqrt{C})^2\ge 0$.

“$\Longrightarrow$”: $\quad$ Clearly, $A \ge 0$ and $C \ge 0$.

If $B < - \sqrt{4AC}$ and $AC > 0$, then $Ax_0^2 + Bx_0 + C < 0$ where $x_0 = \sqrt{\frac{C}{A}} > 0$. Contradiction.

If $B < - \sqrt{4AC}$ and $AC = 0$, there exists $x_1 > 0$ such that $Ax_1^2 + Bx_1 + C < 0$ (easy). Contradiction.

We are done.

$\phantom{2}$

Proof of Fact 2:

By Fact 1, it suffices to prove that $$Ax^2 + Bx + C \ge 0, ~ x\in [0, 1] \iff (A + B + C)s^2 + (B + 2C)s + C \ge 0, ~ s \ge 0.$$

“$\Longleftarrow$”: $\quad$ Clearly, $A+B+C \ge 0$.

If $x = 1$, then $Ax^2 + Bx + C = A + B + C \ge 0$.

If $x\in [0, 1)$, letting $s = \frac{x}{1 - x} \ge 0$, we have $x = \frac{s}{1+s}$ and $$Ax^2 + Bx + C = \frac{1}{(1+s)^2}[(A+B+C)s^2+(B+2C)s+C] \ge 0.$$

“$\Longrightarrow$”: $\quad$ For $s\ge 0$, letting $x = \frac{s}{1+s}\in [0,1]$, we have $$Ax^2+Bx+C = \frac{1}{(1+s)^2}[(A+B+C)s^2+(B+2C)s+C] \ge 0$$ which results in $(A+B+C)s^2+(B+2C)s+C\ge 0$.

We are done.

Answer 2

Since $f$ is of degree 2, its shape it quite simple. Depending on how you want to write your ranges in your answer, I think you were already quite close.

If $f$ is an increasing function on the domain $[0,1]$, then we only have to check your conditions 1 and 2 (try to visualize this). The same is true if $f$ is an decreasing function on the domain $[0,1]$.

The only exception is when $f$ changes from increasing to decreasing (or vice versa) in the interval $[0,1]$. This happens if $0 \leq \dfrac{-b}{2a} \leq 1$, this is the vertex you mention.
In this case we also need that $0 \leq -\dfrac{b^2-4ac}{2a} \leq 1$.

Answer 3

You have $0\le f(x) \le 1$ on $[0,1]$ if and only if $f(0)$, $f(1) \in [0,1]$ and $(\ -\frac{b}{2a} \in [0,1] \implies -\frac{\Delta}{4 a} \in [0, 1]\ )$, that is all of the conditions below are satisfied

1. $ c\in [0,1]$

2. $a + b + c \in [0,1]$

3. $ - \frac{b}{2 a } \not \in [0,1]$ or $-\frac{\Delta}{4 a} \in [0,1]$