Math Problem Statement

Give the definiton of the image of a function f : [a, b] → R, and determine the image of the function f : [−1, 2] → R defined by f (x) = x2 − 4x + 1

Solution

Definition: Image of a Function

The image of a function f:[a,b]Rf : [a, b] \to \mathbb{R} is the set of all output values (range) that the function produces when applied to every input xx in the domain [a,b][a, b]. Mathematically, the image of a function ff is written as: Image(f)={f(x)x[a,b]}.\text{Image}(f) = \{ f(x) \mid x \in [a, b] \}.

Finding the Image of the Function f(x)=x24x+1f(x) = x^2 - 4x + 1

We are given the function f(x)=x24x+1f(x) = x^2 - 4x + 1 and the domain [1,2][-1, 2]. Let's determine its image:

  1. Rewrite the function: The function f(x)=x24x+1f(x) = x^2 - 4x + 1 is a quadratic function. It can be rewritten in vertex form by completing the square: f(x)=(x24x)+1=((x2)24)+1=(x2)23.f(x) = (x^2 - 4x) + 1 = \left( (x - 2)^2 - 4 \right) + 1 = (x - 2)^2 - 3. So, f(x)=(x2)23f(x) = (x - 2)^2 - 3.

  2. Analyze the behavior on the domain [1,2][-1, 2]:

    • The vertex of the quadratic function is at x=2x = 2, and the parabola opens upwards because the coefficient of x2x^2 is positive.
    • At x=2x = 2, the function attains its minimum value: f(2)=(22)23=3.f(2) = (2 - 2)^2 - 3 = -3.
    • Now, evaluate the function at the endpoints of the domain:
      • At x=1x = -1: f(1)=(1)24(1)+1=1+4+1=6.f(-1) = (-1)^2 - 4(-1) + 1 = 1 + 4 + 1 = 6.
      • At x=2x = 2: f(2)=3(as calculated earlier).f(2) = -3 \quad \text{(as calculated earlier)}.
  3. Determine the range (image): Since the quadratic function is continuous and achieves its minimum value at x=2x = 2, and the values at the endpoints are 3-3 and 66, the image of the function f(x)f(x) over the interval [1,2][-1, 2] is the interval: Image(f)=[3,6].\text{Image}(f) = [-3, 6].

Conclusion:

The image of the function f(x)=x24x+1f(x) = x^2 - 4x + 1 on the domain [1,2][-1, 2] is the interval [3,6][-3, 6].

Would you like more details on how the image is derived or have any questions?

Here are some related questions to consider:

  1. How do you complete the square for quadratic functions?
  2. How does the vertex form of a quadratic function help in determining its minimum or maximum?
  3. What are the properties of quadratic functions that determine the shape of their graphs?
  4. Can a quadratic function ever have an unbounded image on a closed interval?
  5. How do we compute the image of more complex functions?

Tip: When determining the image of a function, always check the behavior at critical points (like the vertex in a quadratic) and endpoints of the domain.

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Math Problem Analysis

Mathematical Concepts

Functions
Quadratic Functions
Image of a Function
Completing the Square

Formulas

f(x) = ax^2 + bx + c
Vertex form of a quadratic function: f(x) = (x - h)^2 + k
Image(f) = { f(x) | x ∈ [a, b] }

Theorems

Fundamental Theorem of Algebra (quadratic polynomials)
Properties of Quadratic Functions (vertex form, range)

Suitable Grade Level

Grades 10-12