#Help
47 messages · Page 1 of 1 (latest)
I found the first and second derivate
I’m not sure how to solve for X when setting equal to 0
What did you get?
show the work you currently have so far
its better if you type in the answers you believe is true into the boxes before screenshotting
if you want more general help that doesnt address what youre stuck on, theres a video on doing this: https://youtu.be/15awMHeP1Yc
fyi you can use Win + Shift + W or Cmd + Ctrl + Shift + 4 to screenshot the screen instead of taking a picture on your phone
the command save the part of the screen you want into your clipboard, then you can Ctrl + V that here
its fast and more "proper" that way
This calculus video tutorial shows you how to find the intervals where the function is increasing and decreasing, the critical points or critical numbers, relative extrema such as local minimum and local maximum values using the first derivative test, concavity, and inflection points using the second derivative.
Introduction to Limits: ...
I dont have discord on my laptop
But i know i have to set the first derivative equal to 0
The problem im having is solving for x
I have the first and second derivate
@shy drum
Well no id have to set the first derivative equal to 0
To find increasing or decreasing
And i cant remember how to solve for X for exponentials
And i know they have to be positive so im not sure how that changes things
whats your first derivative then
2e^2x+e^-x
thast not correct
How so
you can use chain rule on e^(2x)
the outside is e^x
the inside is 2x
so e^(2x) * 2
Oh 2e
so the derivative of e^(2x) is 2 e^x
dont say 2e
thats a misnomer
say 2 e instead
also use the chain rule on e^(-x)
-e^x
-e^(-x)
Right
being bold and being careless dont mix well
Mybad i glanced at it im ar work right now
"no" you say?
But still the problem im having is solving for X
you should put some more weight on your Nos next time
then take a second glance at this
the bottom line has one x
then you can "isolate for x" since there is only one x in the equation
that means you just do operations to repeatedly move stuff from the LHS to the RHS
moving e^ to the RHS is done by ln() both sides
use these hints to work out the value of x
I see
.solved