Let's solve the problem **step by step**: ### Step 1: Understand the function Given: \[ f(x) = \begin{cases} 1 - x^2 & \text{if } x \leq 1 \\ \ln(x) & \text{if } x > 1 \end{cases} \] We need to show that \( f \) is continuous on \( (-\infty, \infty) \). ### Step 2: Check continuity on intervals - For \( x < 1 \), \( f(x) = 1 - x^2 \) (a polynomial, always continuous). - For \( x > 1 \), \( f(x) = \ln(x) \) (logarithmic, continuous for \( x > 0 \)). ### Step 3: Check continuity at \( x = 1 \) We need to check: - \( \lim_{x \to 1^-} f(x) \) - \( \lim_{x \to 1^+} f(x) \) - \( f(1) \) #### Left-hand limit (\( x \to 1^- \)): \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (1 - x^2) = 1 - (1)^2 = 0 \] #### Right-hand limit (\( x \to 1^+ \)): \[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \ln(x) = \ln(1) = 0 \] #### Value at \( x = 1 \): \[ f(1) = 1 - (1)^2 = 0 \] ### Step 4: Fill in the answer boxes 1. **First box (left-hand limit):** \( 1 - x^2 \) 2. **Second box (right-hand limit):** \( \ln(x) \) 3. **Third box (limit at \( x \to 1 \)):** \( 0 \) ### Step 5: Final Answer - The left-hand limit: \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (1 - x^2) = 0 \) - The right-hand limit: \( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \ln(x) = 0 \) - So, \( \lim_{x \to 1} f(x) = 0 \) - \( f(1) = 0 \) Thus, \( f \) is continuous at \( x = 1 \), and therefore on \( (-\infty, \infty) \). --- ### **Filled Boxes:** 1. **First box:** \( 1 - x^2 \) 2. **Second box:** \( \ln(x) \) 3. **Third box:** \( 0 \) --- #### **Summary Table:** | Box Position | Answer | |----------------------|-------------| | First | 1 - x^2 | | Second | ln(x) | | Third | 0 | --- If you enter these, the solution will be complete and correct!

Let's carefully analyze the problem and provide a detailed, step-by-step solution. Based on the image and notes, the problem involves a piecewise function and checking its continuity at the point where the definition changes, i.e., at \( x = 1 \). --- ### Step 1: Understand the given function From the notes and typical problem structure, the function is: \[ f(x) = \begin{cases} 1 - x^2 & \text{if } x \leq 1 \\ \ln(x) & \text{if } x > 1 \end{cases} \] Our goal: **Verify whether \( f(x) \) is continuous at \( x = 1 \).** --- ### Step 2: Check continuity on each interval - For \( x < 1 \), \( f(x) = 1 - x^2 \), which is a polynomial function, and polynomials are continuous everywhere. - For \( x > 1 \), \( f(x) = \ln(x) \), which is continuous for \( x > 0 \). **Thus, the only point to check is \( x = 1 \).** --- ### Step 3: Compute the limits at \( x = 1 \) We need to check: \[ \lim_{x \to 1^-} f(x), \quad \lim_{x \to 1^+} f(x), \quad \text{and} \quad f(1) \] --- ### Step 4: Calculate \( f(1) \) Since the function is defined as \( 1 - x^2 \) for \( x \leq 1 \): \[ f(1) = 1 - (1)^2 = 1 - 1 = 0 \] --- ### Step 5: Calculate the left-hand limit \( \lim_{x \to 1^-} f(x) \) As \( x \to 1^- \), \( f(x) = 1 - x^2 \): \[ \lim_{x \to 1^-} (1 - x^2) = 1 - (1)^2 = 1 - 1 = 0 \] --- ### Step 6: Calculate the right-hand limit \( \lim_{x \to 1^+} f(x) \) As \( x \to 1^+ \), \( f(x) = \ln(x) \): \[ \lim_{x \to 1^+} \ln(x) = \ln(1) = 0 \] --- ### Step 7: Conclusion about continuity at \( x=1 \) Since: \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 0 \] the function \( f \) is **continuous at \( x=1 \)**. --- ### **Answer boxes filling:** - **First box** (left-hand limit): **\( 1 - x^2 \)** - **Second box** (right-hand limit): **\( \ln(x) \)** - **Third box** (limit at \( x=1 \)): **\( 0 \)** --- ### **Final note:** Since the function is continuous on both sides and at \( x=1 \), **\( f \) is continuous everywhere**. --- ### **Summary:** | Step | Expression | Explanation | |---------|----------------|--------------| | \( \lim_{x \to 1^-} f(x) \) | \( 1 - x^2 \) | Polynomial, continuous for \( x \leq 1 \) | | \( \lim_{x \to 1^+} f(x) \) | \( \ln(x) \) | Logarithm, continuous for \( x > 0 \) | | \( f(1) \) | \( 1 - 1^2 = 0 \) | Function value at \( x=1 \) | --- ### **Final answer:** - **First box:** \( 1 - x^2 \) - **Second box:** \( \ln(x) \) - **Third box:** \( 0 \) This completes the detailed solution.