https://file.aiquickdraw.com/mathbot/file/1726835583777g7ywrwq6.jpg

Let's analyze the given function and evaluate the provided statements one by one.

### Given function \( f(x) \):
\[
f(x) = 
\begin{cases} 
\sqrt{-x + 882}, & \text{for } x \leq 441 \\
\sqrt{x}, & \text{for } x > 441 
\end{cases}
\]
We are asked to evaluate whether the following statements are true or false:

---

### 1. \( f(x) \geq 1 \) for all \( x \in \mathbb{R} \).

- For \( x \leq 441 \):  
  The expression is \( f(x) = \sqrt{-x + 882} \).  
  The minimum value of \( f(x) \) occurs at \( x = 441 \), where \( f(441) = \sqrt{441} = 21 \). Thus, for \( x \leq 441 \), \( f(x) \geq 21 \), which implies that \( f(x) \geq 1 \).

- For \( x > 441 \):  
  The function is \( f(x) = \sqrt{x} \), and \( \sqrt{x} \geq \sqrt{441} = 21 \). Again, \( f(x) \geq 21 \), so \( f(x) \geq 1 \) holds for this case as well.

**Conclusion**: True.  
---

### 2. There exists \( x > 441 \) such that \( f(x) \leq 20 \).

- For \( x > 441 \), the function is \( f(x) = \sqrt{x} \).  
  Since \( \sqrt{x} > \sqrt{441} = 21 \) for \( x > 441 \), there is no value of \( x > 441 \) where \( f(x) \leq 20 \).

**Conclusion**: False.  
---

### 3. \( \lim_{x \to 441^+} f(x) \neq \lim_{x \to 441^-} f(x) \).

- As \( x \to 441^- \), \( f(x) = \sqrt{-x + 882} \), and \( \lim_{x \to 441^-} f(x) = \sqrt{-441 + 882} = \sqrt{441} = 21 \).
- As \( x \to 441^+ \), \( f(x) = \sqrt{x} \), and \( \lim_{x \to 441^+} f(x) = \sqrt{441} = 21 \).

Since both limits are equal to 21, the statement is false.

**Conclusion**: False.  
---

### 4. \( f \) is continuous at \( x = 441 \).

- Continuity requires that \( \lim_{x \to 441^-} f(x) = \lim_{x \to 441^+} f(x) = f(441) \).  
  We've already calculated that \( \lim_{x \to 441^-} f(x) = 21 \) and \( \lim_{x \to 441^+} f(x) = 21 \).  
  Also, \( f(441) = \sqrt{441} = 21 \).

Since the limits match the function value, \( f \) is continuous at \( x = 441 \).

**Conclusion**: True.  
---

### 5. The domain of \( f \), \( D_f \), is \( \mathbb{R} \).

- For \( x \leq 441 \), we need \( -x + 882 \geq 0 \), which simplifies to \( x \leq 882 \).  
  Thus, \( f(x) \) is defined for all \( x \leq 441 \).
- For \( x > 441 \), \( f(x) = \sqrt{x} \), which is defined for all \( x > 441 \).

Hence, \( D_f = (-\infty, 441] \cup (441, \infty) = \mathbb{R} \).

**Conclusion**: True.  
---

### 6. \( f : D_f \to [21, \infty) \) is surjective.

- For \( x \leq 441 \), \( f(x) = \sqrt{-x + 882} \), and the range of this part is from 21 to 0. However, since the square root is only real for non-negative values, the minimum value is 21.
- For \( x > 441 \), \( f(x) = \sqrt{x} \), and the range is from 21 to infinity.

Thus, the overall range of the function is \( [21, \infty) \), and \( f \) is surjective onto this interval.

**Conclusion**: True.

---

Would you like further details or explanations? Here are some related questions to explore:

1. How do you calculate the limits from both sides of a discontinuity?
2. What conditions must be satisfied for a function to be continuous at a point?
3. How do you determine if a function is surjective?
4. What is the process for finding the domain of a piecewise function?
5. How does the square root function behave over different domains?

**Tip**: When analyzing a piecewise function, always check the boundary points separately for continuity, differentiability, or limits to avoid missing subtle details.

Given the function 
\( f(x) = 
\begin{cases} 
\sqrt{-x + 882}, & \text{for } x \leq 441 \\
\sqrt{x}, & \text{for } x > 441 
\end{cases}
\)
Assess the following statements:
1. \( f(x) \geq 1 \) for all \( x \in \mathbb{R} \).
2. There exists \( x > 441 \) such that \( f(x) \leq 20 \).
3. \( \lim_{x \to 441^+} f(x) \neq \lim_{x \to 441^-} f(x) \).
4. \( f \) is continuous at \( x = 441 \).
5. The domain of \( f \), \( D_f \), is \( \mathbb{R} \).
6. \( f : D_f \to [21, \infty) \) is surjective.

Analyze the given piecewise function with square root components. Learn how to verify continuity, limits, and surjectivity. This problem is suitable for advanced high school students (Grades 11-12) studying calculus concepts.

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