Let f (x) = min { x, $x^2$ } for every real number of x, then
- h is continuous for all x
- h is differentiable for all x
- h ' (x) = 1, $\forall$ x > 1
- h is not differentiable at two values of x
The Correct Option is D
Solution and Explanation
h (x) is continuous all x, but hix) is not differentiable at
two points x = 0 and x = l . (due to sharp edges). Also
h ' (x) = 1, $\forall$ x > 1
Hence, (a), (c) and (d) is correct answers,
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