Limits, Continuity & Differentiability
Differentiability
Grade 12
Question:
<p>If \(f(x) = x^3 \text{sgn}(x)\), then</p>
<p>(a) \(f\) is differentiable at \(x = 0\)</p>
<p>(b) \(f\) is continuous but not differentiable at \(x = 0\)</p>
<p>(c) \(f'(0) = 1\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: The sign function $\text{sgn}(x)$ combined with $x^3$ creates a smooth function at the origin because $x^3$ vanishes at $x=0$, making both left and right derivatives zero.
<p>$f(x) = \begin{cases} -x^3 & x < 0 \\ 0 & x = 0 \\ x^3 & x > 0 \end{cases}$</p><p><strong>Continuity at $x = 0$:</strong> $\lim_{x \to 0^-} f(x) = 0$, $\lim_{x \to 0^+} f(x) = 0$, $f(0) = 0$. Function is continuous.</p><p><strong>Differentiability at $x = 0$:</strong> Left derivative: $\lim_{h \to 0^-} \frac{-h^3}{h} = \lim_{h \to 0^-} (-h^2) = 0$. Right derivative: $\lim_{h \to 0^+} \frac{h^3}{h} = \lim_{h \to 0^+} h^2 = 0$. Both equal 0, so $f'(0) = 0$ and $f$ is differentiable at $x = 0$.</p><p>∴ Answer is (a).</p>
Correct Answer: a