I have been trying for the past hour or so to figure out the solution to this equation:
\sqrt{x} + 1 = 0
but to no avail! Here's what i tried...
\sqrt{x} = -1
\sqrt{x} = i^4
x = \pm i^8
x = \pm 1
But this is most obviously wrong. What did I do wrong and what is the real...
No, and my apologies if my input is therefore irrelevant! I was just saying that that is normal and I don't think any technique other than repetition/relearning will solve it ( i really don't know anything about the technique you mentioned though.)
Unfortunately that is pretty much just how memory works. Though after learning something 2 or 3 times it tends to stick longer. Use it or lose it, basically.
Adopting a frame moving at c would cause dividing-by-zero type absurdities. The first of which being that space would contract 100%. How can you be moving at c if there's nowhere to go? xD
b is an arbitrary real number as well, he's asking why you reverse the sign of b and not a, for which i do not have an answer. my best guess is that its just more useful to define it that way.
On the "left" of 1, the function is defined as F(x)=x^2 +1, then F'(1)=_____
On the "right" of 1, the function is defined as F(x)=2x, then F'(1)=_____
If these two values exist and are equal, then the function is differentiable at 1.
For a function to be differentiable at a point, its left sided and right sided derivatives must be equal. Usually the function is the same on both sides of a point, but for this function it has two different pieces.