Limits, Continuity & Differentiability
Continuity of piecewise functions
Grade 12
Question:
<p>Function \(f(x) = \begin{cases} |x+1| & x < -2 \\ 2x + 3 & -2 \leq x < 0 \\ x^2 + 3 & 0 \leq x < 3 \\ x^3 - 15 & x \geq 3 \end{cases}\)</p><p>(a) is continuous at all points in \(\mathbb{R}\)</p><p>(b) [option cut off in source text]</p>
<p>(a) is continuous at all points in \(\mathbb{R}\)</p>
Step-by-Step Solution
Key Concept: Check continuity at boundary points by comparing left and right limits with the function value.
<p>Check continuity at transition points:</p><p>At $x = -2$: $\lim_{x \to -2^-} |x+1| = |-2+1| = 1$ and $\lim_{x \to -2^+} (2x+3) = -4+3 = -1$. Discontinuous at $x = -2$.</p><p>Since there is a discontinuity at $x = -2$, the function is not continuous at all points in $\mathbb{R}$.</p>
Correct Answer: A