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Let $f$ be a real-valued function of a real variable defined as $f(x) = x^{2}$ for $x\geq0$ and $f(x) = -x^{2}$ for  $x < 0$.Which one of the following statements is true?

  1. $f(x) \text{is discontinuous at  x = 0}$ 
  2. $f(x) \text{is continuous but not differentiable at x = 0} $
  3. $f(x) \text{is differentiable but its first derivative is not continuous at  x = 0 }$
  4. $f(x) \text{is differentiable but its first derivative is not differentiable at x = 0} $
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Ans B)
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option D) is correct.

refer answer below.
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1
ok..
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D is the answer?????
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1 Answer

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option D) is correct.

$f(x)=x^2 , x>=0$

      $=-x^2 , x<0$

at $x=0$ function f(x) is continuous since $LHL=RHL=f(0) = 0$

now, first derivative of f(x)

$f'(x) = 2x , x>=0$

       $=-2x , x<0$

since, $f'(x=0^+)=0$ and $f'(x=0^-)=0$ i.e both are equal,

function f(x) is diffrentiable at $x=0$

derivative of first derivative,

$f''(x) = 2 , x>=0$

      $ =-2 , x<0$

since $f''(x=0^+) = 2$ and $f''(x=0^-) = -2$ i.e both are unequal

derivative of f(x) is not diffrentiable at $x=0$

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