1
$\begingroup$

I know that we should check critical numbers (points where f'(x) is either zero or not defined) and endpoints (for a closed interval) as possible points of local extrema of f(x). Obviously, all these points should be in the domain of f. So why does the following problem also test x = 0, which is not in the domain of f?

enter image description here

$\endgroup$
1
  • $\begingroup$ How come do you claim they test $0$? $\endgroup$
    – Git Gud
    Nov 22, 2014 at 14:36

1 Answer 1

0
$\begingroup$

In order for the testing to work, the function must be continuous on the entire test interval. This function is not continuous (or even defined) at $x=0$. As you can see, the function in fact changes direction as one moves from the left to the right of 0.

$\endgroup$
2
  • $\begingroup$ So the solution is wrong? $\endgroup$
    – Leponzo
    Nov 22, 2014 at 15:29
  • $\begingroup$ @Leponzo No, it's correct. This is the reason the interval needs to be split at 0. $\endgroup$ Nov 22, 2014 at 15:36

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.