let f(x) = (x-8)^2 find a domain in which f is one to one and non decreasing. then find the inverse of f restricted to this domain

The given function is \( f(x) = (x - 8)^2 \).

### Step 1: Identify when \( f(x) \) is one-to-one and non-decreasing.
To ensure that a function is one-to-one and non-decreasing, its derivative must be non-negative (\( f'(x) \geq 0 \)).

#### Derivative of \( f(x) \):
\[
f'(x) = 2(x - 8)
\]
The derivative \( f'(x) = 2(x - 8) \) is non-negative when \( x \geq 8 \).

Thus, \( f(x) \) is non-decreasing for \( x \geq 8 \). We will restrict the domain to \( x \geq 8 \) to make \( f(x) \) one-to-one and non-decreasing.

### Step 2: Restrict the domain of \( f(x) \).
The domain of \( f(x) \) where it is one-to-one and non-decreasing is \( [8, \infty) \).

### Step 3: Find the inverse of \( f(x) \) on this restricted domain.
To find the inverse of \( f(x) \), follow these steps:

1. Start with \( y = (x - 8)^2 \).
2. Solve for \( x \) in terms of \( y \):
   \[
   \sqrt{y} = x - 8
   \]
   \[
   x = \sqrt{y} + 8
   \]
   
   Since we are considering the domain \( x \geq 8 \), we only take the positive square root.

### Final Answer:
- **Domain**: \( [8, \infty) \)
- **Inverse function**: \( f^{-1}(x) = \sqrt{x} + 8 \), for \( x \geq 0 \).

Would you like further details or have any questions?

Here are some related questions to explore:
1. What does it mean for a function to be one-to-one?
2. How does restricting a domain affect the invertibility of a function?
3. Why did we choose the positive square root when finding the inverse?
4. What are the conditions for a function to have an inverse?
5. How would the inverse change if the function had a different form?

**Tip**: When solving for an inverse, always check if the original function needs domain restrictions to ensure it is one-to-one.

Explore how to find a domain in which the function f(x) = (x-8)^2 is one-to-one and non-decreasing. Learn to determine the inverse of f(x) on this restricted domain using derivatives and inverse function rules. Suitable for high school students in Grades 10-12.

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