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Let's work through the questions in the image one by one:

### Given: 
\[ f(x) = x^2 - 6x + 5 \]

---

### (i) **Express \( f(x) \) in the form \( (x - a)^2 - b \):**

To express \( f(x) = x^2 - 6x + 5 \) in the form \( (x - a)^2 - b \), we need to **complete the square**:

1. Start with \( f(x) = x^2 - 6x + 5 \).
2. Take half the coefficient of \(x\), which is \(-6\). Half of \(-6\) is \(-3\), and squaring it gives \(9\).
3. Add and subtract \(9\) inside the equation to complete the square:
   \[
   f(x) = (x^2 - 6x + 9) - 9 + 5
   \]
4. Simplify:
   \[
   f(x) = (x - 3)^2 - 4
   \]

Thus, the function in the required form is:
\[
f(x) = (x - 3)^2 - 4
\]

---

### (ii) **Hence, state the range of the function \(f(x)\):**

Since \(f(x)\) is written as \(f(x) = (x - 3)^2 - 4\), and \((x - 3)^2 \geq 0\) for all real \(x\), the minimum value of \((x - 3)^2\) is 0, which occurs when \(x = 3\).

Therefore, the minimum value of \(f(x)\) is:
\[
f(3) = 0 - 4 = -4
\]
The function can take any value greater than or equal to \(-4\).

**Range:**
\[
f(x) \geq -4
\]
Or in interval notation:
\[
\text{Range} = [-4, \infty)
\]

---

### (iii) **State the largest possible domain so that the function \(f(x)\) is increasing:**

The function \(f(x) = (x - 3)^2 - 4\) is a quadratic function with a minimum at \(x = 3\). To ensure the function is increasing, we need to restrict the domain to the values of \(x\) where the derivative of \(f(x)\) is positive.

The function starts increasing after the vertex, \(x = 3\). So the largest possible domain where the function is increasing is:
\[
x \geq 3
\]

**Domain:**
\[
[3, \infty)
\]

---

### (iv) **Find the inverse function \(f^{-1}(x)\), for the 1-1 function in part (iii), stating its domain:**

To find the inverse, we start with \(f(x) = (x - 3)^2 - 4\), but we restrict \(f(x)\) to the domain \(x \geq 3\) so that it is 1-1.

1. Set \(y = f(x)\):
   \[
   y = (x - 3)^2 - 4
   \]
2. Solve for \(x\) in terms of \(y\):
   \[
   y + 4 = (x - 3)^2
   \]
   \[
   x - 3 = \sqrt{y + 4}
   \]
   \[
   x = 3 + \sqrt{y + 4}
   \]

Thus, the inverse function is:
\[
f^{-1}(x) = 3 + \sqrt{x + 4}
\]

**Domain of the inverse:**
Since the expression inside the square root must be non-negative, \(x + 4 \geq 0\), which gives \(x \geq -4\).

Therefore, the domain of \(f^{-1}(x)\) is:
\[
x \geq -4
\]
Or in interval notation:
\[
\text{Domain of } f^{-1}(x): [-4, \infty)
\]

---

Would you like further clarification on any part of this? Here are 5 related questions you could explore:

1. How do you verify if a function is 1-1?
2. What is the process of completing the square for any quadratic function?
3. How do you find the inverse of more complex functions?
4. What is the geometric interpretation of the range and domain of a function?
5. How does restricting the domain of a quadratic function affect its inverse?

**Tip:** To check if a quadratic function is 1-1, consider its graph's behavior. It will only be 1-1 on a restricted domain where it is either increasing or decreasing consistently.

A function f is defined by f(x) = x^2 - 6x + 5, x ∈ R. (i) Express f(x) in the form (x - a)^2 - b. (ii) Hence state the range of the function f(x). (iii) State the largest possible domain so that the function f(x) is increasing. (iv) Find the inverse function f^{-1}(x), for the 1-1 function in part (iii), stating its domain.

Learn how to express the quadratic function f(x) = x^2 - 6x + 5 in completed square form, find its range, determine the largest possible increasing domain, and calculate the inverse function. This problem is suitable for high school students studying advanced algebra.

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