Limits, Continuity & Differentiability
Differentiability
Grade 12
Question:
<p>Let \(f: R \to R\) be differentiable at \(c \in R\) and \(f(c) = 0\). If \(g(x) = |f(x)|\), then at \(x = c\), \(g\) is</p>
<p>Not differentiable if \(f'(x) = 0\).</p>
<p>Differentiable if \(f'(c) \neq 0\).</p>
<p>Differentiable if \(f'(c) = 0\).</p>
<p>Not differentiable.</p>
Step-by-Step Solution
Key Concept: For g(x) = |f(x)| to be differentiable at c, we need f'(c) = 0 when f(c) = 0. If f'(c) ≠ 0, the graph has a sharp corner at c making |f(x)| non-differentiable there.
<p><strong>Step 1:</strong> Since f(c) = 0 and f is differentiable at c, consider the definition of differentiability of g(x) = |f(x)| at x = c.</p><p><strong>Step 2:</strong> The left derivative: $g'_-(c) = \lim_{h \to 0^-} \frac{|f(c+h)| - |f(c)|}{h} = \lim_{h \to 0^-} \frac{|f(c+h)|}{h}$</p><p><strong>Step 3:</strong> Using f(c) = 0 and expanding f(c+h) = f'(c)·h + o(h): $g'_-(c) = \lim_{h \to 0^-} \frac{|f'(c)·h + o(h)|}{h}$</p><p><strong>Step 4:</strong> Similarly, right derivative: $g'_+(c) = \lim_{h \to 0^+} \frac{|f'(c)·h + o(h)|}{h}$</p><p><strong>Step 5:</strong> For both limits to exist and be equal, we need the left and right limits of $\frac{|f'(c)·h|}{h}$ to match. This requires f'(c) = 0.</p><p><strong>Step 6:</strong> If f'(c) ≠ 0, the sign of f(c+h) changes differently on left and right, creating a corner (non-differentiable). If f'(c) = 0, then g'(c) = 0 (differentiable).</p><p><strong>Conclusion:</strong> g(x) is continuous at x = c (always true since f is continuous), but differentiable only when f'(c) = 0. The general answer is: <strong>g is continuous but not necessarily differentiable</strong> (or <strong>differentiable iff f'(c) = 0</strong> if that's option C).</p><p>∴ Answer: C</p>
Correct Answer: C