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

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

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

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

2. **Analyze the behavior on the domain \( [-1, 2] \):**
   - The vertex of the quadratic function is at \( x = 2 \), and the parabola opens upwards because the coefficient of \( x^2 \) is positive.
   - At \( x = 2 \), the function attains its minimum value:
     \[
     f(2) = (2 - 2)^2 - 3 = -3.
     \]
   - Now, evaluate the function at the endpoints of the domain:
     - At \( x = -1 \):
       \[
       f(-1) = (-1)^2 - 4(-1) + 1 = 1 + 4 + 1 = 6.
       \]
     - At \( x = 2 \):
       \[
       f(2) = -3 \quad \text{(as calculated earlier)}.
       \]

3. **Determine the range (image):**
   Since the quadratic function is continuous and achieves its minimum value at \( x = 2 \), and the values at the endpoints are \( -3 \) and \( 6 \), the image of the function \( f(x) \) over the interval \( [-1, 2] \) is the interval:
   \[
   \text{Image}(f) = [-3, 6].
   \]

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

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.

Give the definition 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) = x^2 − 4x + 1

Discover the image of the quadratic function f(x) = x^2 - 4x + 1 on the domain [-1, 2] with a step-by-step solution. This explanation involves key concepts such as completing the square, analyzing the vertex, and determining the range of the function. Suitable for high school students in Grades 10-12.

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