It is a proven and taught **fact** that a function increasing or decreasing is based on the sign of its derivative

Here's proof that you're incorrect. The FACT is that the derivative

**is not present in the definition of an increasing function whatsoever**. I've even provided a counter-example

**of a function with a discontinuous derivative which is always increasing**. The Wolfram article even says this, giving

**another example** of a function that is always increasing despite having a function that doesn't have a defined derivative at all points.

I can prove it right now in this example:

-snip-

This is an example function. And I’m finding where it increases/decreases. First you have to solve for when the derivative of the function is 0 or undefined. I plot these values on a number line and I use the principle of continuity to find out whether the derivative is positive or negative. Near those values, the **+** means increasing, the **-** means decreasing

Let’s see if I’m correct:

-snip-

There’s the actual function.

And now we see that your method is wrong - the domain for which your function is increasing INCLUDES the point x=+1. Firstly, f(+1)=1/(1-2-3)=-1/4. -1/4 is

**BIGGER THAN OR EQUAL TO** every other point on the interval x in (-1,1) - so, we would include it in the interval, so the function is ACTUALLY increasing on the interval x in (-1,1]. You are correct that the derivative can be used (and is often easier to do so) to find when a function is increasing, but because the definition IS NOT f'(x)>0, you need to be careful when applying it.

That’s because it’s a point of inflexion. Which is where the second derivative is equal to 0.

What's that gotta do with the price of fish in China? It's not even clear which point of S_R_K's you're trying to argue with here. What you've said is correct, but that doesn't mean ANYTHING that they've said is incorrect.

You have missed my point completely. It doesn’t even matter the domain. The instantaneous velocity will still be positive at t=0

Do you realize that the velocity of an objects displacement is equal to the first derivative of the function of said displacement?

No, you are missing the point here. The domain DOES matter, because it determines where a derivative is defined. Again,

**as S_R_K told you before**, this function is not defined for t<0 - as a consequence, this means that

**the function's derivative CANNOT be defined for t=0**, because the left-hand limit (or the limit from the negative side)

**cannot** be defined. Remember, even though we have a bunch of fancy rules for derivatives, the fundamental truth of the derivative is:

If that limit does not exist, then the derivative does not exist. And in this case:

Since f(0+h) is NOT defined for h<0, then the limit CANNOT exist.