Limits, Continuity & Differentiability
Differentiability of piecewise functions
Grade 12
Question:
<p>If a function \(f(x)\) is defined as
\[f(x) = \begin{cases}
-x & , x \leq 0 \\
x^2 & , 0 < x \leq 1 \\
x^2 - x + 1 & , x > 1
\end{cases}\]
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 differentiable at \(x = 1\) but not at \(x = 0\)</p>
<p>(d) \(f(x)\) is not differentiable at \(x = 0\) and \(x = 1\)</p>
Step-by-Step Solution
Key Concept: Check left and right derivatives at boundary points; if they differ, the function is not differentiable at that point.
<p><strong>Solution:</strong> For differentiability at $x = 0$:</p>
<p>Left derivative: $L f'(0^-) = \lim_{h \to 0^-} \frac{f(0-h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-(-h) - 0}{-h} = -1$</p>
<p>Right derivative: $R f'(0^+) = \lim_{h \to 0^+} \frac{h^2 - 0}{h} = 0$</p>
<p>Since $L f'(0^-) \neq R f'(0^+)$, the function is not differentiable at $x = 0$.</p>
<p>For differentiability at $x = 1$:</p>
<p>Left derivative at $x = 1$: $f'(1^-) = 2(1) = 2$</p>
<p>Right derivative at $x = 1$: $f'(1^+) = 2(1) - 1 = 1$</p>
<p>Since the derivatives do not match, the function is not differentiable at $x = 1$.</p>
<p>∴ Answer is (d) $f(x)$ is not differentiable at $x = 0$ and $x = 1$</p>
Correct Answer: D