Question

What now? Calculus @zepdrix

Answer 1

@zepdrix

Answer 2

@StudyGurl14
hey this the same thing
except that there are not extrema

Answer 3

Critical points also exist where x is undefined, btw. Not just where g' = 0

Answer 4

g'<0 is still valid for this problem
because e^-x^2 >0 positive for any x
with minus sign it becomes <0

Answer 5

yes and i said that in your earlier post

Answer 6

How do you know it will be positive for any x?

Answer 7

Oh, sorry. Didn't catch that bit.

Answer 8

but your function so far are all differentiable at any x

Answer 9

well think of e^-x^2
plug in any x value and it will turn a positive for the output

Answer 10

y=e^-x^2
y values are always positive

Answer 11

okay.

Answer 12

in fact e^x as a graph is above x axis
that won't change for e^-x^2

Answer 13

So there are no extrema?>