Given the function $f:\mathbb{R}\to \mathbb{R}$. defined as $$f(x)=e^{|x|+x^2}+|x^2-1|$$ Which of the following is true about the function $f$.

  1. It is not differentiable exactly at three points of $\mathbb{R}$.
  2. It is not differentiable at $x=0$.
  3. It is differentiable at $x=2$.
  4. It is not differentiable at $x=1$ and $x=-1$.

So what i did is consider $f$ as sum of two functions one involving the exponential term and other $|x^2-1|$ and try to prove the differentiability of the two functions on the indicated points separately and then by using the fact that sum of two differentiable function is differentiable, conclude that $f$ is differentiable or not. I succeeded in showing $2$ and $4$ true by this method but couldn't able to reach any conclusion about $3$. Help me out?

  • 1
    $\begingroup$ Hint: Remove modulus because everything is positive at x=2 and now check for differentiability $\endgroup$ – Sujith Sizon Oct 15 '15 at 13:51

Hint: In a small neighborhood of $x=2$, you have $$f(x)=e^{x+x^2}+(x^2-1).$$ This a sum of differentiable functions, hence is differentiable.

  • $\begingroup$ just curiosity what happen if we take a small neighborhood around $x=1$. Can we conclude not differentiability at $x=1$ by this line of thinking? $\endgroup$ – Kushal Bhuyan Oct 15 '15 at 13:58
  • $\begingroup$ @KprimeX: Correct; in a small neighborhood of $x=1$, $|x|=x$, thus the first term in the sum is differentiable. If $f(x)$ were differentiable, then $f(x)-e^{x+x^2}$ would be differentaible (i.e., $|x^2-1|$ would be differentiable). This will allow you to work with the "problem" term. $\endgroup$ – Clayton Oct 15 '15 at 16:38
  • $\begingroup$ So, all are true. right? $\endgroup$ – Math geek Jun 9 at 17:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.