tmagloire1
  • tmagloire1
ap calc ab help please ! http://prntscr.com/8o63v0 http://prntscr.com/8o640n
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
f' is the line of the slope of f
anonymous
  • anonymous
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'
tmagloire1
  • tmagloire1
How can I find that without an equation?

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anonymous
  • anonymous
You can look at how line A decreases on the interval -infinity to zero
anonymous
  • anonymous
and then you can see that line b is negative from -infinity to zero
tmagloire1
  • tmagloire1
This problem is so confusing omg. Okay so is line B decreasing from .4- -infinit?
anonymous
  • anonymous
Yeah something like that
anonymous
  • anonymous
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
tmagloire1
  • tmagloire1
I think line B is positive and increasing while line A is decreasing
anonymous
  • anonymous
|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
anonymous
  • anonymous
I'd just tell you the answer but I think its kinda important to understand. Give me a sec
tmagloire1
  • tmagloire1
ok thanks
anonymous
  • anonymous
Okay so I marked where line B is increasing / decreasing
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anonymous
  • anonymous
You can see that line b is negative while a is decreasing
anonymous
  • anonymous
as well as positive while a is increasing
tmagloire1
  • tmagloire1
Yep so that would mean that it's f normal right
anonymous
  • anonymous
since the derivative is the line of the slope we know b is the derivative
tmagloire1
  • tmagloire1
Ohh ok
anonymous
  • anonymous
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
tmagloire1
  • tmagloire1
Oh ok so you just find where it's decreasing and negative and if the slope is negative than it's the derivative
anonymous
  • anonymous
Yeah but it doesnt have to be decreasing AND negative, just decreasing.
tmagloire1
  • tmagloire1
oh ok i understand thanks for going through the work to help me with that one
anonymous
  • anonymous
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
anonymous
  • anonymous
Basically the number you want to find (the point) is a. So since you're finding 2, a is equal to 2.
tmagloire1
  • tmagloire1
So i would just try plugging them into the two definition of limit equations and see what comes out?
anonymous
  • anonymous
Not really your thing would look like (f(2+h) - f(2))/h
anonymous
  • anonymous
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.
tmagloire1
  • tmagloire1
wait what 2nd part are you referring to
anonymous
  • anonymous
the - f(2)
anonymous
  • anonymous
Do you understand how to plug everything in?
anonymous
  • anonymous
You dont need to actually do it for this problem, but thats important because sometimes you do. Its kinda situational
tmagloire1
  • tmagloire1
im not sure i understand how to plug them in
anonymous
  • anonymous
Alright
anonymous
  • anonymous
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
anonymous
  • anonymous
actually more like ((2+h)^2 + 3(2+h) + 1 - (2)^2 + 3(2) + 1)/h
anonymous
  • anonymous
So then you just do algebra and simplify to get answers
anonymous
  • anonymous
But you can tell this one is c because c has x -> 2 as the limit instead of h -> 0
tmagloire1
  • tmagloire1
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
tmagloire1
  • tmagloire1
@swagmaster47 how can you tell the other answers are correct or not

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