A problem by Ankit Nigam

Level pending

Let h ( x ) = m i n { x , x 2 } h\left( x \right) =min\left\{ x,{ x }^{ 2 } \right\} for every real number of x. Then

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h is not differentiable at two values of x h is not continuous h is differentiable for all values of x none of these

Ankit Nigam by Ankit Nigam , 23, India

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1 solution

Saurabh Patil
Apr 22, 2015

Though the function is continuous for all real numbers but the function is non differentiable at x = 1 , − 1. x = 1 , - 1 .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Correction: the two points are x = 0 and 1 considering:

h(x) = x (for x = (-infinity, 0) U [1, infinity) ); x^2 (for x = [0, 1) );

tom engelsman - 3 years, 2 months ago

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