So I've been doing derivatives for a while now, and I was always taught that functions with removable discontinuities, like\(f(x) = \frac{x+1}{x+1}\cdot x^2\) , are not differentiable, because at the removed point (in this case \(f(-1)\)), the derivative is undefined. I even went online and searched this question to confirm it to be true.

However, I was going through https://mualphatheta.org/content/past-tests-2021 (the Mu individual test, question 3) and the solution used the definition of a derivative as \(\lim_{h \to 0} \frac{f(x+5h)-f(x-4h)}{9h}\) instead of the usual \(\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\). In other words, this definition approaches the point from both sides simultaneously, instead of one point being on f(x) and the other one approaching from one side, and it seems like a perfectly valid definition.

But doesn't that mean that a function that has a removable discontinuity should be differentiable using this definition? If both sides of the removal discontinuity are continuous, then there shouldn't be any problems finding the derivative at that point, right?

thanks in advance for anyone who can figure this out.

textot Dec 26, 2021

#1**0 **

nvm I think I got it

I think the left and right derivative definition only applies if the normal one works

or else stuff like the derivative with respect to x of |x| at 0 could be set to any slope possible

also for functions with discontinuities, it's just better to find a function that fills that point in.

textot Dec 28, 2021