Question

Let \(h(x) = \min(x, x^2)\), for every real number \(x\). Then,

• A. \(h\) is not continuous for all \(x\)
• B. \(h\) is differentiable for all \(x\)
• C. \(h'(x) \neq 1\) for all \(x > 1\)
• D. \(h\) is not differentiable at two values of \(x\)

Hint:

For min/max functions, check points where the two functions intersect for potential non-differentiability.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Analyze where \(x\) and \(x^2\) intersect.

Step 2: Detailed Explanation:
\(x = x^2 \implies x(x-1)=0 \implies x=0\) or \(x=1\).
For \(x<0\), \(x^2 > x\)? Actually when \(x<0\), \(x^2\) positive, \(x\) negative, so \(h(x)=x\).
For \(0<x<1\), \(x^2 < x\), so \(h(x)=x^2\).
For \(x>1\), \(x^2 > x\), so \(h(x)=x\).
Check differentiability at \(x=0\): left derivative = 1, right derivative = 0. Not differentiable.
At \(x=1\): left derivative = 2, right derivative = 1. Not differentiable.
So \(h\) is not differentiable at two values \(x=0\) and \(x=1\).

Step 3: Final Answer:
Option (D) is correct.