1. **Problem 1:** Determine which statement about the piecewise function \( f(x) = \begin{cases} 2 & x < 5 \\ 2x - 4 & x \geq 5 \end{cases} \) is true regarding differentiability at \( x=5 \). 2. **Check continuity at \( x=5 \):** Calculate left-hand limit: $$\lim_{x \to 5^-} f(x) = 2$$ Calculate right-hand limit: $$\lim_{x \to 5^+} f(x) = 2(5) - 4 = 10 - 4 = 6$$ Since \( 2 \neq 6 \), \( f \) is not continuous at \( x=5 \). 3. **Conclusion for Problem 1:** Since \( f \) is not continuous at \( x=5 \), it cannot be differentiable there. **Answer:** A) \( f \) is not differentiable at \( x=5 \) because \( f \) is not continuous at \( x=5 \). --- 4. **Problem 2:** Determine which statement about the function \( f \) cannot be used to conclude that \( f \) is defined at \( x=1 \). Statements: A) \( \lim_{x \to 1} f(x) \) exists. B) \( f \) is continuous at \( x=1 \). C) \( f \) is differentiable at \( x=1 \). D) The tangent line to \( f \) at \( x=1 \) exists. 5. **Explanation:** - If \( f \) is continuous at \( x=1 \), then \( f(1) \) is defined. - If \( f \) is differentiable at \( x=1 \), then \( f(1) \) is defined. - If the tangent line exists at \( x=1 \), then \( f(1) \) is defined. - However, the existence of \( \lim_{x \to 1} f(x) \) alone does not guarantee \( f(1) \) is defined. **Answer:** A) \( \lim_{x \to 1} f(x) \) exists cannot be used to conclude \( f(1) \) is defined. --- 6. **Problem 3:** Determine which statement about the piecewise function \( f(x) = \begin{cases} x^2 - 20 & x < 5 \\ -x^2 + 20 & x \geq 5 \end{cases} \) is true regarding differentiability at \( x=5 \). 7. **Check continuity at \( x=5 \):** Left-hand limit: $$\lim_{x \to 5^-} f(x) = 5^2 - 20 = 25 - 20 = 5$$ Right-hand limit: $$\lim_{x \to 5^+} f(x) = -5^2 + 20 = -25 + 20 = -5$$ Since \( 5 \neq -5 \), \( f \) is not continuous at \( x=5 \). 8. **Check differentiability:** Since \( f \) is not continuous at \( x=5 \), it is not differentiable there. **Answer:** A) \( f \) is not differentiable at \( x=5 \) because \( f \) is not continuous at \( x=5 \). --- **Summary:** - For the first and third piecewise functions, the function is not continuous at \( x=5 \), so it is not differentiable there. - For the question about \( x=1 \), the existence of the limit alone does not guarantee the function is defined at that point.