Math Problem Statement

Сумата на най-малката и най-голямата стойност на функцията f(x) = 1 - 4x - 2x^2 в интервала [-2, 1] е:

Solution

The problem is to find the sum of the minimum and maximum values of the function f(x)=14x2x2f(x) = 1 - 4x - 2x^2 over the interval [2,1][-2, 1].

Step 1: Analyze the function

The given function is a quadratic function:

f(x)=2x24x+1f(x) = -2x^2 - 4x + 1

The coefficient of x2x^2 is negative (2-2), so the parabola opens downward. This means the function has a maximum value at its vertex, and the minimum values will occur at the endpoints of the interval.

Step 2: Find the critical point

The critical point is where the derivative of f(x)f(x) is zero. Compute the derivative:

f(x)=4x4f'(x) = -4x - 4

Set f(x)=0f'(x) = 0:

4x4=0    x=1-4x - 4 = 0 \implies x = -1

The critical point is x=1x = -1. Check if this point lies in the interval [2,1][-2, 1]. Since x=1x = -1 is within [2,1][-2, 1], it is a valid candidate for the maximum or minimum value.

Step 3: Evaluate f(x)f(x) at the critical point and endpoints

Now calculate f(x)f(x) at x=2x = -2, x=1x = -1, and x=1x = 1:

  1. At x=2x = -2: f(2)=2(2)24(2)+1=2(4)+8+1=8+8+1=1f(-2) = -2(-2)^2 - 4(-2) + 1 = -2(4) + 8 + 1 = -8 + 8 + 1 = 1

  2. At x=1x = -1: f(1)=2(1)24(1)+1=2(1)+4+1=2+4+1=3f(-1) = -2(-1)^2 - 4(-1) + 1 = -2(1) + 4 + 1 = -2 + 4 + 1 = 3

  3. At x=1x = 1: f(1)=2(1)24(1)+1=2(1)4+1=24+1=5f(1) = -2(1)^2 - 4(1) + 1 = -2(1) - 4 + 1 = -2 - 4 + 1 = -5

Step 4: Identify the maximum and minimum values

From the calculations:

  • Maximum value: f(1)=3f(-1) = 3
  • Minimum value: f(1)=5f(1) = -5

Step 5: Compute the sum

The sum of the maximum and minimum values is:

3+(5)=23 + (-5) = -2

Final Answer:

The sum of the maximum and minimum values of f(x)f(x) over the interval [2,1][-2, 1] is:

2\boxed{-2}


Additional Related Questions:

  1. How do we determine whether a quadratic function has a maximum or minimum value?
  2. What does the vertex of a parabola represent in terms of optimization?
  3. How does the interval of evaluation impact the extrema of a function?
  4. Why is it necessary to evaluate endpoints in addition to critical points for optimization?
  5. Can we use a different method (e.g., completing the square) to find the maximum value of f(x)f(x)?

Tip: For optimization problems on a closed interval, always evaluate the function at critical points and the endpoints, as the extrema could occur at any of these points.

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Math Problem Analysis

Mathematical Concepts

Quadratic Functions
Optimization
Calculus

Formulas

Derivative: f'(x) = -4x - 4
Vertex formula for a parabola: x = -b/(2a)

Theorems

Extreme Value Theorem (ensures maximum and minimum values on a closed interval)

Suitable Grade Level

Grades 10-12