Limits, Continuity & Differentiability
Differentiability at Transition Points
Grade 12
Question:
<p>Let \(h(x) = \min\{x; x^2\}\) for every real number \(x\). Then, which one of the following is true?</p>
<p>(a) \(h\) is not continuous for all \(x\)</p>
<p>(b) \(h\) is differentiable for all \(x\)</p>
<p>(c) \(h'(x) = 1\), for all \(x\)</p>
<p>(d) \(h\) is not differentiable at two values of \(x\)</p>
Step-by-Step Solution
Key Concept: The function $h(x) = \min\{x, x^2\}$ changes its definition at the intersection points. Check continuity and differentiability at these transition points.
<p><strong>Step 1:</strong> Find points of intersection of $y = x$ and $y = x^2$: $x = x^2 \Rightarrow x = 0, 1$. Thus, intersection points are $(0, 0)$ and $(1, 1)$.</p><p><strong>Step 2:</strong> Sketch graphs of $y = x$ and $y = x^2$ to identify which function is the minimum in each region.</p><p><strong>Step 3:</strong> For $x \leq 0$: $x \leq x^2$, so $h(x) = x$. For $0 \leq x \leq 1$: $x^2 \leq x$, so $h(x) = x^2$. For $x \geq 1$: $x \leq x^2$, so $h(x) = x$.</p><p><strong>Step 4:</strong> Therefore, $$h(x) = \begin{cases} x & \text{if } x \leq 0 \text{ or } x \geq 1 \\ x^2 & \text{if } 0 \leq x \leq 1 \end{cases}$$</p><p><strong>Step 5:</strong> Check continuity: At $x = 0$: $\lim_{x \to 0^-} h(x) = 0 = h(0) = \lim_{x \to 0^+} h(x)$. At $x = 1$: $\lim_{x \to 1^-} h(x) = 1 = h(1) = \lim_{x \to 1^+} h(x)$. So $h$ is continuous for all $x$.</p><p><strong>Step 6:</strong> Check differentiability: For $x < 0$: $h'(x) = 1$. For $0 < x < 1$: $h'(x) = 2x$. For $x > 1$: $h'(x) = 1$. At $x = 0$: Left derivative = 1, right derivative = 0. At $x = 1$: Left derivative = 2, right derivative = 1. Since left and right derivatives are not equal at $x = 0$ and $x = 1$, $h$ is not differentiable at these two points.</p><p>∴ Answer is (d).</p>
Correct Answer: D