Max Value Of |2ax+b| Given Constraints

Aug 28, 2025 by Benjamin Cohen 39 views

Finding the Maximum Value of |2ax + b| Under Constraints

Hey guys! Let's dive into a cool math problem today. We're going to figure out the maximum possible value of |2ax + b|, given that |x| ≤ 1 and |ax² + bx + c| ≤ 1. This is a fun one that combines algebra and a bit of precalculus thinking. So, grab your thinking caps, and let's get started!

The Problem

Let's break this down step by step and make sure we really understand what's going on. The heart of the problem lies in those inequalities. Understanding the constraints is key. We know that x can be any number between -1 and 1, inclusive. And for any x in that range, the quadratic expression ax² + bx + c must also be between -1 and 1. This gives us a bounded playing field within which we need to find our maximum value for |2ax + b|.

The expression |2ax + b| looks suspiciously like the derivative of our quadratic ax² + bx + c. This is a crucial observation! The derivative connects the rate of change of the quadratic function, and we're trying to maximize that rate, in a way. This hints that the extreme values of the quadratic within our interval will be critical. We can expect that plugging in the boundary values of our interval for x which is -1, 0, and 1, will give us key equations to work with. This is a common strategy in these types of problems – leveraging the given bounds to create a system of equations or inequalities.

Moreover, because we're dealing with absolute values, we need to consider both positive and negative possibilities for the quadratic. Absolute values mean we have to consider symmetry and both positive and negative cases. This often doubles the amount of work but also provides additional clues and constraints. We need to be mindful of this as we move forward, ensuring we don't overlook any potential scenarios. Our goal is to find the largest possible magnitude of 2ax + b, regardless of its sign. Let’s roll up our sleeves and dive into the nitty-gritty details of the solution.

Initial Attempts and Observations

Okay, so let's start with a common approach. We'll plug in some strategic values for x within the given range |x| ≤ 1. The most obvious choices are x = 1, x = -1, and x = 0. These are the boundaries and the midpoint of our interval, and they often give us valuable information.

Plugging in x = 1

If we substitute x = 1 into the inequality |ax² + bx + c| ≤ 1, we get:

|a(1)² + b(1) + c| ≤ 1
|a + b + c| ≤ 1

Let's call this equation (*). This tells us that the absolute value of the sum a + b + c is less than or equal to 1. This is a crucial piece of information, so let's keep it handy.

Plugging in x = -1

Now, let's try plugging in x = -1:

|a(-1)² + b(-1) + c| ≤ 1
|a - b + c| ≤ 1

We'll call this equation (**). So, the absolute value of a - b + c is also less than or equal to 1. We're building a system of inequalities here, which is a good sign.

Plugging in x = 0

Finally, let's plug in x = 0:

|a(0)² + b(0) + c| ≤ 1
|c| ≤ 1

We'll label this as equation (***). This one is pretty straightforward: the absolute value of c is less than or equal to 1. This gives us a direct bound on the value of c.

Now we have three equations:

(*) |a + b + c| ≤ 1
(**) |a - b + c| ≤ 1
(***) |c| ≤ 1

These inequalities are the foundation of our solution. They provide constraints on the relationships between a, b, and c. But how do we use these to find the maximum value of |2ax + b|? This is where things get interesting.

Our next challenge is to somehow manipulate these inequalities to isolate |2ax + b| or at least find an upper bound for it. We need to think creatively about how to combine these equations to eliminate variables or create expressions that involve 2a and b directly. Remember, we're aiming for the maximum value, so we're looking for the tightest possible bound. We know that the expression 2ax + b resembles the derivative of the given quadratic. This could be a clue to use the inequalities to bound the coefficients of the quadratic.

Strategic Manipulation of Inequalities

Okay, guys, let's get strategic! We've got our three key inequalities, and now we need to figure out how to combine them to get closer to our goal: finding the maximum value of |2ax + b|. The key here is clever manipulation. We want to somehow isolate or bound the expression 2ax + b.

Adding and Subtracting Equations

Let’s start by playing around with equations (*) and (**). If we add them, something interesting happens:

|a + b + c| + |a - b + c| ≤ 1 + 1
|a + b + c| + |a - b + c| ≤ 2

This looks promising! Notice how the b terms have opposite signs in the two absolute values. Adding the inequalities might help us eliminate b or at least reduce its influence. But directly adding absolute values isn't very helpful, we should instead consider specific values for x.

What if we subtract equation (**) from equation (*)?

|(a + b + c) - (a - b + c)| ≤ |a + b + c| + |a - b + c| ≤ 1 + 1
|2b| ≤ 2
|b| ≤ 1

Boom! We've got a bound on |b|. This is a significant step forward. We now know that the absolute value of b is less than or equal to 1. This gives us one piece of the puzzle. But we still need to figure out how a fits into the picture. Let’s keep this in our back pocket and explore further.

Let's add equations (*) and (**), after removing the absolute values, we have:

-1 <= a + b + c <= 1
-1 <= a - b + c <= 1

Adding these inequalities gives us:

-2 <= 2a + 2c <= 2
-1 <= a + c <= 1

We already know that |c| ≤ 1, so let’s use this to try and isolate a. We can write -1 ≤ c ≤ 1. Then from -1 ≤ a + c ≤ 1, we can try to eliminate c. We want to find bounds on a as well, but this seems less straightforward than finding the bound on b. So, what else can we do?

Thinking About the Target Expression

Let’s take a step back and remind ourselves what we’re trying to maximize: |2ax + b|. We already have a bound on |b|, which is great. But we need to deal with the 2ax term. The x here is a variable within the range [-1, 1]. So, 2ax can vary depending on the value of x. This means we might need to consider different values of x to find the maximum of the entire expression.

Perhaps we can try expressing 2ax + b as a linear combination of the expressions we already have bounds for (a + b + c, a - b + c, and c). This is a common technique in optimization problems: try to express the target function in terms of the constrained functions. If we can do that, we can use the bounds on those expressions to find a bound on our target.

This might involve some clever algebraic manipulation. We need to figure out how to write 2ax + b as a sum of multiples of a + b + c, a - b + c, and c. This might seem tricky, but let’s give it a shot. We want to find coefficients p, q, and r such that:

2ax + b = p(a + b + c) + q(a - b + c) + r(c)

If we can find such p, q, and r, then we can use the triangle inequality to bound |2ax + b|:

|2ax + b| = |p(a + b + c) + q(a - b + c) + r(c)|
|2ax + b| ≤ |p||a + b + c| + |q||a - b + c| + |r||c|
|2ax + b| ≤ |p| * 1 + |q| * 1 + |r| * 1
|2ax + b| ≤ |p| + |q| + |r|

Our goal now is to find those coefficients p, q, and r. This is the next step in our adventure! Let's keep pushing forward, guys.

Finding the Coefficients and the Final Bound

Alright, let's crack this coefficient problem! We're trying to express 2ax + b as a linear combination of (a + b + c), (a - b + c), and c. This means we need to find p, q, and r such that:

2ax + b = p(a + b + c) + q(a - b + c) + r(c)

To find these coefficients, we'll expand the right side and then equate the coefficients of a, b, and c on both sides. This will give us a system of equations that we can solve.

Expanding the right side, we get:

2ax + b = pa + pb + pc + qa - qb + qc + rc
2ax + b = (p + q)a + (p - q)b + (p + q + r)c

Now, let's equate the coefficients:

Coefficient of a: 2x = p + q
Coefficient of b: 1 = p - q
Coefficient of c: 0 = p + q + r

We now have a system of three equations with three unknowns. We can solve this using standard techniques.

Solving the System of Equations

Let's solve this system. From the first two equations, we have:

p + q = 2x
p - q = 1

Adding these equations, we get:

2p = 2x + 1
p = x + 1/2

Subtracting the second equation from the first, we get:

2q = 2x - 1
q = x - 1/2

Now, we can substitute p and q into the third equation:

0 = p + q + r
0 = (x + 1/2) + (x - 1/2) + r
0 = 2x + r
r = -2x

So, we've found our coefficients:

p = x + 1/2
q = x - 1/2
r = -2x

Bounding |2ax + b|

Now we can use these coefficients to bound |2ax + b|:

|2ax + b| = |p(a + b + c) + q(a - b + c) + r(c)|
|2ax + b| ≤ |p||a + b + c| + |q||a - b + c| + |r||c|
|2ax + b| ≤ |x + 1/2| * 1 + |x - 1/2| * 1 + |-2x| * 1
|2ax + b| ≤ |x + 1/2| + |x - 1/2| + 2|x|

We're getting closer! Now we need to maximize this expression over the interval [-1, 1]. This involves considering different intervals for x based on the absolute values.

Maximizing the Expression

Let's analyze the expression |x + 1/2| + |x - 1/2| + 2|x|. We'll consider different cases for x:

Case 1: -1 ≤ x ≤ -1/2

In this case, x + 1/2 ≤ 0, x - 1/2 < 0, and x < 0. So,
```
|x + 1/2| = -(x + 1/2)
|x - 1/2| = -(x - 1/2)
|x| = -x
```
Thus, the expression becomes:
```
-(x + 1/2) - (x - 1/2) - 2x = -x - 1/2 - x + 1/2 - 2x = -4x
```
Since -1 ≤ x ≤ -1/2, the maximum value in this interval occurs at x = -1/2, and the value is -4(-1/2) = 2.
Case 2: -1/2 ≤ x ≤ 0

In this case, x + 1/2 ≥ 0, x - 1/2 < 0, and x ≤ 0. So,
```
|x + 1/2| = x + 1/2
|x - 1/2| = -(x - 1/2)
|x| = -x
```
The expression becomes:
```
x + 1/2 - (x - 1/2) - 2x = x + 1/2 - x + 1/2 - 2x = 1 - 2x
```
Since -1/2 ≤ x ≤ 0, the maximum value in this interval occurs at x = -1/2, and the value is 1 - 2(-1/2) = 2.
Case 3: 0 ≤ x ≤ 1/2

In this case, x + 1/2 > 0, x - 1/2 ≤ 0, and x ≥ 0. So,
```
|x + 1/2| = x + 1/2
|x - 1/2| = -(x - 1/2)
|x| = x
```
The expression becomes:
```
x + 1/2 - (x - 1/2) + 2x = x + 1/2 - x + 1/2 + 2x = 1 + 2x
```
Since 0 ≤ x ≤ 1/2, the maximum value in this interval occurs at x = 1/2, and the value is 1 + 2(1/2) = 2.
Case 4: 1/2 ≤ x ≤ 1

In this case, x + 1/2 > 0, x - 1/2 ≥ 0, and x > 0. So,
```
|x + 1/2| = x + 1/2
|x - 1/2| = x - 1/2
|x| = x
```
The expression becomes:
```
x + 1/2 + x - 1/2 + 2x = 4x
```
Since 1/2 ≤ x ≤ 1, the maximum value in this interval occurs at x = 1, and the value is 4(1) = 4.

The Grand Finale

Comparing the maximum values from all cases, we see that the maximum value of |x + 1/2| + |x - 1/2| + 2|x| is 4.

Therefore, the maximum possible value of |2ax + b| is 4.

Conclusion

So, there you have it! We've successfully navigated this intricate problem and found that the maximum possible value of |2ax + b| is 4. This journey involved some clever algebraic manipulation, strategic substitutions, and careful analysis of absolute values. We saw how breaking down the problem into smaller steps and considering different cases can lead us to the solution.

This type of problem is a great exercise in mathematical thinking and problem-solving. It challenges us to use our knowledge of inequalities, absolute values, and algebraic techniques in a creative way. Keep practicing, guys, and you'll become math problem-solving ninjas in no time! Remember, the key is to understand the constraints, manipulate the given information, and think strategically about the target expression. You've got this!