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ISI2014-DCG-29

0 votes
149 views

If $f(x) = \sin \bigg( \dfrac{1}{x^2+1} \bigg),$ then

  1. $f(x)$ is continuous at $x=0$, but not differentiable at $x=0$
  2. $f(x)$ is differentiable at $x=0$, and $f’(0) \neq 0$
  3. $f(x)$ is differentiable at $x=0$, and $f’(0) = 0$
  4. None of the above
in Calculus
recategorized by 149 views

1 Answer

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Here, $\sin\left(\frac{1}{x^{2}+1}\right)$ is continuous and definitely differentiable.

$f(x)=\sin\left(\frac{1}{x^{2}+1}\right)\\ \Rightarrow f'(x)= \frac{-2x}{\left(x^{2}+1\right)^2}\cos\left(\frac{1}{x^{2}+1}\right)\\ \Rightarrow f'(0)=0$

 

So the correct answer is C.

 

0

Here's the graph of $\sin\left(\frac{1}{x^{2}+1}\right)$ below.

 

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