Math Problem Statement

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.

Solution

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.

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Math Problem Analysis

Mathematical Concepts

Piecewise Functions
Continuity
Limits
Surjectivity
Function Domain

Formulas

Square root function properties
Limit properties

Theorems

Continuity at a point
Limit theorems
Properties of the square root function

Suitable Grade Level

Grades 11-12

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