Question:

<p><strong>Step 1: Check continuity at x = 1</strong></p><p>For continuity at x = 1, we need: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$</p><p><strong>Left-hand limit:</strong> Using the first piece since $x \leq 1$:<br/>$\lim_{x \to 1^-} f(x) = \frac{1}{4}(3(1)^2 + 1) = \frac{1}{4}(4) = 1$</p><p><strong>Value at x = 1:</strong> Since x = 1 falls in the first piece ($-\infty < x \leq 1$):<br/>$f(1) = \frac{1}{4}(3(1)^2 + 1) = 1$</p><p><strong>Right-hand limit:</strong> Using the second piece since $1 < x < 4$:<br/>$\lim_{x \to 1^+} f(x) = 5 - 4(1) = 1$</p><p>All three are equal to 1, so $f$ is <strong>continuous at x = 1</strong>. ✓</p><p><strong>Step 2: Check continuity at x = 4</strong></p><p>For continuity at x = 4, we need: $\lim_{x \to 4^-} f(x) = \lim_{x \to 4^+} f(x) = f(4)$</p><p><strong>Left-hand limit:</strong> Using the second piece since $1 < x < 4$:<br/>$\lim_{x \to 4^-} f(x) = 5 - 4(4) = 5 - 16 = -11$</p><p><strong>Right-hand limit:</strong> Using the third piece since $4 < x < \infty$:<br/>$\lim_{x \to 4^+} f(x) = 4 - 4 = 0$</p><p><strong>Issue with f(4):</strong> The point x = 4 is NOT included in any piece of the function (first piece ends at x ≤ 1, second piece is $1 < x < 4$, third piece is $4 < x < \infty$). However, for the limit definition of continuity, f(4) need not be explicitly defined. The critical issue is that the left and right limits are not equal:</p><p>$\lim_{x \to 4^-} f(x) = -11 \neq 0 = \lim_{x \to 4^+} f(x)$</p><p>Therefore, $f$ is <strong>discontinuous at x = 4</strong>. ✗</p><p><strong>Conclusion:</strong> The function is continuous at x = 1 and discontinuous at x = 4.</p><p>$\therefore$ <strong>Answer: C</strong></p>