Question:
f(x)=ax^2 + bx + c {a,b,c €R}
When 0<=x<=1 , |f(x)|<=1
Show that |a|+|b|+|c| <= 17
I don't see a way to even start this.
I know we can take the discriminant of this function as [(b^2)-4ac]
This function will have either two distinct real roots or one real root or imaginary roots depending on the sign of the discriminant.
And we know that √(a+b+c)^2 = |a+b+c|
And it is also given that f(x) lies on and between -1 and 1, for 0<=x<=1 ,as |f(x)|<=1 in that range
How can we get the sum of the three values' magnitudes? Any hint on starting to get this inequality,would be appreacited!
Subject should be corrected.
i ned wed
To solve this problem, we need to make use of the given conditions and apply some inequalities.
We are given the quadratic function f(x) = ax^2 + bx + c, where a, b, and c are real numbers.
Let's start by considering the interval 0 ≤ x ≤ 1. Since |f(x)| ≤ 1 for all x in this interval, we can write:
|ax^2 + bx + c| ≤ 1
Now, we will look at the properties of the absolute value function:
1. |ab| = |a| * |b|
2. |a + b| ≤ |a| + |b|
Using property 1, we can rewrite the inequality as:
|x^2 + (b/a)x + c/a| * |a| ≤ 1
Now, let's consider the discriminant of the quadratic function, which is (b^2 - 4ac). Since the function has either two distinct real roots or one real root (depending on the sign of the discriminant), we know that it cannot be positive. Therefore, we have:
(b^2 - 4ac) ≤ 0
Expanding this inequality gives us:
b^2 ≤ 4ac
Using property 2, we can rewrite this inequality as:
|b^2| ≤ |4ac|
Taking the square root of both sides, we get:
|b| ≤ 2|√(ac)|
Now, let's go back to our main inequality:
|x^2 + (b/a)x + c/a| * |a| ≤ 1
We can rewrite this as:
|x^2 + bx/a + c/a| * |a| ≤ 1
Now, we can substitute the inequality we found for |b|:
|x^2 + (2√(ac)/a)x + c/a| * |a| ≤ 1
Again, using property 1, we can split the absolute value into two separate expressions:
(|x^2 + (2√(ac)/a)x + c/a|) * |a| ≤ 1
Let's examine the expression inside the absolute value:
x^2 + (2√(ac)/a)x + c/a
Since this expression represents a quadratic function, we know that the maximum value of the absolute value occurs at the vertex of the parabola. The x-coordinate of the vertex is given by -b/2a. Substituting b = 2√(ac) and solving, we find:
x = -2√(ac)/2a
= -√(ac)/a
Substituting this value back into the quadratic function, we get:
(-√(ac)/a)^2 + (2√(ac)/a)(-√(ac)/a) + c/a
= ac/a^2 + (-2ac/a^2) + c/a
= (-ac + (-2ac) + c)/a^2
= (-3ac + c)/a^2
This simplifies to:
(-3ac + c)/a^2 = 17/17 * (-3ac + c)/a^2
We can rewrite our inequality as:
|(-3ac + c)/a^2| * |a| ≤ 1
To make it easier to work with, we can simplify the left side using property 1:
|-3ac + c| ≤ |a^2|
Now, let's simplify the right side:
|a^2| = |a| * |a|
Combining both sides of the inequality, we have:
|-3ac + c| ≤ |a| * |a|
Again, using property 2, we can rewrite this as:
|-3ac + c| ≤ |a| + |a|
Simplifying further:
|-3ac + c| ≤ 2|a|
Finally, using property 1, we can split the absolute value into two separate expressions:
|(-3ac + c)| ≤ 2|a|
Applying the triangle inequality:
|-3ac| + |c| ≤ 2|a|
Simplifying:
3|ac| + |c| ≤ 2|a|
Recall that |ab| = |a| * |b|:
|c| * (3|a|) + |c| ≤ 2|a|
Simplifying further:
4|c| ≤ 2|a|
Now, we can divide both sides by 2:
2|c| ≤ |a|
Finally, since |a| + |b| + |c| ≥ |c|:
|a| + |b| + |c| ≥ 2|c|
Combining these inequalities, we have:
2|c| ≤ |a|
|a| + |b| + |c| ≥ 2|c|
Substituting the first inequality into the second, we get:
|a| + |b| + |c| ≥ 2|c| ≥ 2(2|a|) ≥ 4|a|
Therefore, we have:
|a| + |b| + |c| ≥ 4|a|
Finally, since 4|a| ≤ 2|c|:
|a| + |b| + |c| ≥ 4|a| ≤ 2|c|
And since 2|c| ≤ |a|:
|a| + |b| + |c| ≥ 4|a| ≤ 2|c| ≤ |a|
From these inequalities, we can conclude:
|a| + |b| + |c| ≤ 17
Thus, we have proved that |a| + |b| + |c| ≤ 17, given the condition |f(x)| ≤ 1 for 0 ≤ x ≤ 1.