Limits, Continuity & Differentiability
Differentiability
Grade 12
Question:
<p>If a function \(f(x)\) is defined as \(f(x) = \begin{cases} 7x & x < 0 \\ 2 & 0 \leq x \leq 1 \\ 2 + 7x - 2x^2 & x > 1 \end{cases}\), then</p>
<p>(a) \(f(x)\) is differentiable at \(x = 0\) and \(x = 1\)</p>
<p>(b) \(f(x)\) is differentiable at \(x = 0\) but not at \(x = 1\)</p>
<p>(c) \(f(x)\) is not differentiable at \(x = 1\) but differentiable at \(x = 0\)</p>
<p>(d) \(f(x)\) is not differentiable at \(x = 0\) and \(x = 1\)</p>
Step-by-Step Solution
Key Concept: A function is differentiable at a point if and only if the left-hand and right-hand derivatives exist and are equal. Check both points separately.
<p><strong>Analysis at $x = 0$:</strong> Left derivative: $\lim_{h \to 0^-} \frac{7h - 0}{h} = 7$. Right derivative: $\lim_{h \to 0^+} \frac{2 - 2}{h} = 0$. Since left and right derivatives differ, $f$ is not differentiable at $x = 0$.</p><p><strong>Analysis at $x = 1$:</strong> Left derivative from constant region: $0$. Right derivative: $\lim_{h \to 0^+} \frac{(2 + 7(1+h) - 2(1+h)^2) - 2}{h} = \lim_{h \to 0^+} \frac{7h - 4h - 2h^2}{h} = 3$. Since left and right derivatives differ, $f$ is not differentiable at $x = 1$.</p><p>∴ Answer is (d).</p>
Correct Answer: d