Math Problem Statement

f(x)= x^2-6x+8; a=1, b=3

Solution

Given the function f(x)=x26x+8f(x) = x^2 - 6x + 8, we are interested in analyzing the behavior of the function over the interval from a=1a = 1 to b=3b = 3.

Let's go step by step:

  1. Find the value of f(x)f(x) at the endpoints:
    • f(1)=(1)26(1)+8=16+8=3f(1) = (1)^2 - 6(1) + 8 = 1 - 6 + 8 = 3
    • f(3)=(3)26(3)+8=918+8=1f(3) = (3)^2 - 6(3) + 8 = 9 - 18 + 8 = -1

So, f(1)=3f(1) = 3 and f(3)=1f(3) = -1.

  1. Find the derivative of f(x)f(x) to check for critical points: The derivative of f(x)f(x) is: f(x)=2x6f'(x) = 2x - 6 Setting the derivative equal to zero to find critical points: 2x6=0x=32x - 6 = 0 \quad \Rightarrow \quad x = 3

    The critical point occurs at x=3x = 3, but since this is an endpoint of the interval, we don't have any additional critical points inside the interval [1,3][1, 3].

  2. Analyze the behavior of f(x)f(x) over the interval:

    • At x=1x = 1, the function value is 3.
    • At x=3x = 3, the function value is -1.

    The function decreases from f(1)=3f(1) = 3 to f(3)=1f(3) = -1.

Would you like to explore more details on this or perform further calculations?

Here are some follow-up questions:

  1. What is the vertex of the parabola represented by f(x)f(x)?
  2. How does the function behave beyond the interval [1,3][1, 3]?
  3. What are the roots of the function f(x)f(x)?
  4. Can you rewrite f(x)f(x) in vertex form?
  5. How can we determine if the function has a maximum or minimum within the interval?

Tip: Finding the vertex of a quadratic function helps you understand where the function reaches its maximum or minimum value!

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Math Problem Analysis

Mathematical Concepts

Quadratic Functions
Derivatives
Critical Points
Interval Analysis

Formulas

Quadratic function: f(x) = ax^2 + bx + c
Derivative of a quadratic function: f'(x) = 2ax + b

Theorems

Critical point occurs where the derivative is zero
Quadratic functions have a vertex which represents the maximum or minimum value

Suitable Grade Level

Grades 9-11