Understanding the Function and Domain

The given function is \(f(x) = x^2 - 4x + 5\). This is a quadratic function, which represents a parabola. The domain of the function is specified as \(A = (1, 4)\). This is an open interval, meaning \(x\) can take any value between 1 and 4, but not including 1 or 4.

To find the range of the function over this specific domain, we need to understand the behavior of the parabola and its values within the interval \(x \in (1, 4)\).

Analyzing the Parabola's Vertex

A quadratic function of the form \(f(x) = ax^2 + bx + c\) has a vertex at \(x = -\frac{b}{2a}\). For our function \(f(x) = x^2 - 4x + 5\), we have \(a=1\), \(b=-4\), and \(c=5\). The x-coordinate of the vertex is:

\(x = -\frac{-4}{2(1)} = \frac{4}{2} = 2\)

The y-coordinate (the function's value) at the vertex is:

\(f(2) = (2)^2 - 4(2) + 5 = 4 - 8 + 5 = 1\)

The vertex of the parabola is at \((2, 1)\).

Checking the Vertex Location Relative to the Domain

The domain is the open interval \((1, 4)\). The x-coordinate of the vertex is 2. We need to check if this vertex lies within the given domain:

\(1 < 2 < 4\)

Yes, the vertex at \(x=2\) is within the domain \((1, 4)\). Since the coefficient \(a=1\) is positive, the parabola opens upwards. This means the vertex represents the minimum point of the parabola. Because the vertex is inside the domain \((1, 4)\), the minimum value of the function within this domain is the value at the vertex, which is \(f(2) = 1\). This minimum value of 1 is included in the range because \(x=2\) is included in the domain \((1, 4)\).

Evaluating the Function at the Domain Endpoints

Although the domain is an open interval and the endpoints \(x=1\) and \(x=4\) are not included, evaluating the function at these points helps us determine the behavior of the function as \(x\) approaches these values. The function's value approaches \(f(1)\) as \(x \to 1^+\) and approaches \(f(4)\) as \(x \to 4^-\).

• At \(x=1\): \(f(1) = (1)^2 - 4(1) + 5 = 1 - 4 + 5 = 2\)
• At \(x=4\): \(f(4) = (4)^2 - 4(4) + 5 = 16 - 16 + 5 = 5\)

As \(x\) goes from 1 towards 2, the function value decreases from a value close to 2 down to the minimum value 1. As \(x\) goes from 2 towards 4, the function value increases from 1 towards a value close to 5.

Determining the Range of the Function

The range consists of all possible values \(f(x)\) can take for \(x \in (1, 4)\). We found the minimum value within the domain is \(f(2) = 1\). Since \(x=2\) is in the domain, the value 1 is in the range. The function increases as \(x\) moves away from the vertex within the domain. The function approaches \(f(1)=2\) as \(x \to 1^+\) and approaches \(f(4)=5\) as \(x \to 4^-\).

The values \(f(x)\) range from the minimum value up to the maximum value reached within the interval. The maximum value occurs at one of the endpoints of the effective interval based on distance from the vertex. The points 1 and 4 are equally distant from the vertex \(x=2\) (both are 2 units away). The function value increases as we move away from the vertex \(x=2\). So, the function values go up towards 2 (as \(x \to 1^+\)) and up towards 5 (as \(x \to 4^-\)). The highest value approached is 5.

Since the interval is open \((1, 4)\), the function values \(f(1)=2\) and \(f(4)=5\) are not strictly included in the range. However, the function takes all values between its minimum (1) and the supremum of the values approached at the boundaries (5). The minimum value is 1, which is attained at \(x=2\), so 1 is included. The function values go up towards, but do not reach, 5 as \(x\) approaches 4. Therefore, the range starts from 1 (included) and goes up to 5 (excluded).

The range is \([1, 5)\).

Summary of Range Calculation Steps

1. Identify the function and the given domain.
2. For a quadratic function, find the vertex.
3. Check if the vertex x-coordinate is within the domain.
4. If the vertex is within the domain, the y-coordinate of the vertex is either the minimum (if parabola opens up) or maximum (if parabola opens down) value within that domain, and it is included in the range.
5. Evaluate the function at the endpoints of the domain interval.
6. Determine the range based on the vertex value and the values approached at the endpoints. For an open interval, the endpoint values are typically excluded from the range unless they are also attained at the vertex.

Based on the minimum value at the vertex (1, included) and the values approached at the endpoints (up to 5, excluded), the range of the function \(f(x) = x^2 - 4x + 5\) for \(x \in (1, 4)\) is \([1, 5)\).

Revision Table: Key Concepts

Additional Information: Range of Quadratic Functions on Intervals

Finding the range of a quadratic function \(f(x) = ax^2 + bx + c\) over a given interval \([x_1, x_2]\) or \((x_1, x_2)\) or mixed involves considering the vertex and the function values at the interval's endpoints.

• If the vertex x-coordinate (\(x_v\)) is within the interval \((x_1, x_2)\): The range will include the vertex y-coordinate (\(f(x_v)\)) as either the minimum (if \(a>0\)) or maximum (if \(a<0\)) value. The other bound of the range will be the maximum or minimum of \(f(x_1)\) and \(f(x_2)\), depending on the direction of the parabola and the interval type (open or closed).
• If the vertex x-coordinate (\(x_v\)) is outside the interval \([x_1, x_2]\): The function is monotonic (either strictly increasing or strictly decreasing) over the interval. The range is then determined solely by the function values at the endpoints, \(f(x_1)\) and \(f(x_2)\). The range will be [\(\min(f(x_1), f(x_2))\), \(\max(f(x_1), f(x_2))\)]. The inclusion/exclusion of the endpoints in the range depends on whether the interval \([x_1, x_2]\) is open or closed at those ends.
• For open intervals like \((x_1, x_2)\), even if an endpoint value \(f(x_1)\) or \(f(x_2)\) is the maximum or minimum approached, it will be excluded from the range unless that same value is attained at a point *within* the open interval (like at the vertex).

In our specific case, the vertex \(x=2\) is within the open interval \((1, 4)\), and \(f(2)=1\) is the minimum. The function increases as \(x\) moves towards 1 and 4. The values approached are \(f(1)=2\) and \(f(4)=5\). The maximum value approached is 5. Since 1 is the minimum and it's included, and 5 is the maximum value approached but not reached, the range is \([1, 5)\).