Limits, Continuity & Differentiability
Continuity
Grade 12
Question:
<p>Consider the piecewise defined function $f(x) = \begin{cases} -x & ; x < 0 \\ 0 & ; 0 \leq x \leq 4 \\ x - 4 & ; x > 4 \end{cases}$. Choose the answer which best describes the continuity of this function.</p>
<p>(a) the function is unbounded and therefore cannot be continuous.</p>
<p>(b) the function is right continuous at $x = 0$.</p>
<p>(c) the function has a removable discontinuity at 0 and 4, but is continuous on the rest of the real line.</p>
<p>(d) the function is continuous on the entire real line.</p>
Step-by-Step Solution
Key Concept: For piecewise functions, check continuity at transition points by verifying that LHL = RHL = function value. Also verify each piece is continuous within its domain.
<p><strong>Step 1:</strong> Check continuity at $x = 0$:</p><p>$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x) = 0$</p><p>$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 0 = 0$</p><p>$f(0) = 0$</p><p>All three are equal, so $f$ is continuous at $x = 0$. ✓</p><p><strong>Step 2:</strong> Check continuity at $x = 4$:</p><p>$\lim_{x \to 4^-} f(x) = \lim_{x \to 4^-} 0 = 0$</p><p>$\lim_{x \to 4^+} f(x) = \lim_{x \to 4^+} (x - 4) = 0$</p><p>$f(4) = 0$</p><p>All three are equal, so $f$ is continuous at $x = 4$. ✓</p><p><strong>Step 3:</strong> On each piece, the function is polynomial and hence continuous. Combined with continuity at transition points, $f$ is continuous on $\mathbb{R}$.</p><p>∴ Answer is (d).</p>
Correct Answer: D