12:11 PM
-1

I need to show that $f(x) = \sin\left(\dfrac1x\right)$ at $x \neq 0$ and $f(0) = 0$ satisfies the conclusion of the Intermediate Value Theorem on $[-1,1]$. I need to consider the cases where $a, b < 0; a, b > 0; a < 0 < b; a = 0; b = 0$.

IIRC the tag for intermediate value theorem was discussed (and removed) before. Here it was introduced under the name : chat.stackexchange.com/transcript/3740/2017/3/22 Some past discussions on meta: Do some calculus theorems deserve to have their own tag? and Tag proposal: mean-value-theorem
I would say that result of the most recent discussion on meta was inconclusive.
I have removed this instance of the tag.
@an4s The tag for intermediate value theorem was discussed on meta a few times. See here: Tag proposal: mean-value-theorem and Do some calculus theorems deserve to have their own tag? If you think that such tag might be useful, probably the best thing to do would be discuss the new tag on meta first - either in a separate question or making a new post in the tag management thread. — Martin Sleziak 22 secs ago