ap calc ab help please ! http://prntscr.com/8o63v0 http://prntscr.com/8o640n

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ap calc ab help please ! http://prntscr.com/8o63v0 http://prntscr.com/8o640n

Mathematics
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f' is the line of the slope of f
So on your first problem if you can find which line is positive when one is increasing and negative while the other is decreasing it's f'
How can I find that without an equation?

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Other answers:

You can look at how line A decreases on the interval -infinity to zero
and then you can see that line b is negative from -infinity to zero
This problem is so confusing omg. Okay so is line B decreasing from .4- -infinit?
Yeah something like that
Really you just eyeball where one is negative and see if the other line is decreasing, and if the other is positive one is increasing
I think line B is positive and increasing while line A is decreasing
|dw:1444085317486:dw| you can see that the straight line is the derivative because where its positive the arch is increasing and where its negative its decreasing
I'd just tell you the answer but I think its kinda important to understand. Give me a sec
ok thanks
Okay so I marked where line B is increasing / decreasing
1 Attachment
You can see that line b is negative while a is decreasing
as well as positive while a is increasing
Yep so that would mean that it's f normal right
since the derivative is the line of the slope we know b is the derivative
Ohh ok
Do you get it? A is decreasing so the slope is negative, the slope of the line is the derivative. Look for the line that is negative while decreasing
Oh ok so you just find where it's decreasing and negative and if the slope is negative than it's the derivative
Yeah but it doesnt have to be decreasing AND negative, just decreasing.
oh ok i understand thanks for going through the work to help me with that one
Your next problem is about the definition of a derivative at a point which is defined by this http://archives.math.utk.edu/visual.calculus/2/definition.8/eq1.gif
Basically the number you want to find (the point) is a. So since you're finding 2, a is equal to 2.
So i would just try plugging them into the two definition of limit equations and see what comes out?
Not really your thing would look like (f(2+h) - f(2))/h
So for f(2+h) you plug (2+h) wherever you see x and f(2) where ever you see h for the 2nd part.
wait what 2nd part are you referring to
the - f(2)
Do you understand how to plug everything in?
You dont need to actually do it for this problem, but thats important because sometimes you do. Its kinda situational
im not sure i understand how to plug them in
Alright
So I have (f(2+h) - f(2))/h My equation is x^2 + 3x + 1 f(2+h) would be (2+h)^2 + 3(2+h) + 1 f(2) would be 2^2 + 3(2) + 1 I now have ((2+h)^2 + 3(2+h) + 1 - 2^2 + 3(2) + 1)/h
actually more like ((2+h)^2 + 3(2+h) + 1 - (2)^2 + 3(2) + 1)/h
So then you just do algebra and simplify to get answers
But you can tell this one is c because c has x -> 2 as the limit instead of h -> 0
oh ok so you take ((2+h)^2 + 3(2+h) + 1 - (2)^2 + 3(2) + 1)/h and simplify to see if i get the other answers
@swagmaster47 how can you tell the other answers are correct or not

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