If $|ax^2 + bx + c| \le 1$ for all $x$ in $[0,1]$ then evaluate $|a|$, $|b|$ and $|c|$

Question

Solve the following:

If $|ax^2 + bx + c| \le 1$ for all $x$ in $[0,1]$, then
i) $|a| \le 8$
ii) $|b| \ge 8$
iii) $|c| \le 1$
iv) $|a| + |b| + |c| \le 17$

Solution from my textbook:

Put $x = 0, 1 \text{ and } \frac{1} {2}$ to get: $$|c| \le 1$$ $$|a + b + c| \le 1$$ $$|a + 2b + 4c| \le 4$$ From the above three equations, we get, $|b| \le 8 \text{ and } |a| \le 8$. Therefore, $|a| + |b| + |c| \le 17$

However, I don't understand: how, from the three equations, can you find the values of $|a|$ and $|b|$?

Answer 1

$-1\leq c\leq 1,\ -1\leq a+b+c\leq 1,\ -4\leq a+2b+4c\leq 4$

$-5\leq b+3c\leq\ 5,\ -8\leq b\leq 8$

$-3\leq a+b+3c\leq 3,\ -8\leq a\leq8$

$\therefore |a|+|b|+|c|\leq 17$

Each time you add or subtract, the left and right ranges increase. This is a problem of adding or subtracting as few times as possible from the narrowest possible range.

Answer 2

Using triangle inequality to simplify the other answer, we have for example:

$$|b+3c|=|a+2b+4c-(a+b+c)|\le|a+2b+4c|+|a+b+c|\le4+1=5$$

$$|b|=|b+3c-3c|\le |b+3c|+|3c|\le5+3=8$$

Answer 3

Let $f(x)=ax^2+bx+c$.

Thus, $$a+b+c=f(1),$$ $$\frac{1}{4}a+\frac{1}{2}b+c=f\left(\frac{1}{2}\right)$$ and $$c=f(0),$$ which gives $$b=4f\left(\frac{1}{2}\right)-f(1)-3f(0),$$ $$a=2f(1)+2f(0)-4f\left(\frac{1}{2}\right)$$ and by the triangle inequality we obtain: $$|a|+|b|+|c|\leq2+2+4+4+1+3+1=17.$$ It's interesting that for $f(x)=8x^2-8x+1$ we have an equality.