max of min and min of max type

also i didnt go too conceptual in this but you could just take put in values and see, like if go for negative values x could go to - infinty and then be the minimum value. thus we need positive values, for x=1, x is the minimum but 4-x is also pretty close i.e. 3. if somehow we could get middling value between them we could get the answer i.e x=2 and 4-x also gets you two, you go with x=3 and the minimum value of the function is just 1 through 4-x. thus hit and trial could also be your buddy in this

Looking at this optimization problem, I need to find where y = min(x, 2x+1, 4-x) reaches its maximum value.

Let me start by finding the intersection points of these three linear functions:

• f₁(x) = x
• f₂(x) = 2x + 1
• f₃(x) = 4 - x

Finding intersection points:

1. x = 2x + 1 → x = -1

2. x = 4 - x → x = 2

3. 2x + 1 = 4 - x → x = 1

So the critical points are at x = -1, x = 1, and x = 2.

Determining which function is minimum in each interval:

Let me check the values at these intersection points:

• At x = -1: f₁(-1) = -1, f₂(-1) = -1, f₃(-1) = 5
• At x = 1: f₁(1) = 1, f₂(1) = 3, f₃(1) = 3
• At x = 2: f₁(2) = 2, f₂(2) = 5, f₃(2) = 2

Now analyzing each interval:

For x ≤ -1:

Comparing slopes and values, 2x + 1 ≤ x when x ≤ -1, so min = 2x + 1

For -1 ≤ x ≤ 2:

Here x is the smallest function, so min = x

For x ≥ 2:

Here 4 - x becomes smaller than x, so min = 4 - x

Finding the maximum:

• For x ≤ -1: y = 2x + 1 increases as x increases, reaching y = -1 at x = -1
• For -1 ≤ x ≤ 2: y = x increases linearly, reaching y = 2 at x = 2
• For x ≥ 2: y = 4 - x decreases as x increases, starting at y = 2 when x = 2

The function reaches its peak at the boundary point x = 2, where the middle piece (y = x) meets the right piece (y = 4 - x).

Answer:

• Maximum value of y = 2
• Achieved at x = 2