Question

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!

Answer 1

Subject should be corrected.

Answer 2

i ned wed

Answer 3

[AI-generated answer — Explain Bot]

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.