Question

@jim_thompson5910

Calculus fun woo!

http://cyh.leeschools.net/UserContent/Documents/AP%20CalcBC%20SumAssign%2014-15.pdf

Answer 1

Now I don't have to upload every file :P

We left off on 18

Answer 2

That sounds right

Answer 3

have fun

Answer 4

Is it B?

Answer 5

yep that's when g' is negative

Answer 6

Woo I remember because you said if it's the graph of g' and it's under the x axis then it's negative. And those are the only ones fully under the x axis

Answer 7

yeah negative derivative intervals correspond to decreasing intervals

Answer 8

on the original function

Answer 9

I don't know what to do with 19, I don't remember from when we learned that

Answer 10

I have an idea but I'm not sure because I over-think everything

Answer 11

how do we find the slope of a tangent line?

Answer 12

Uhmm the derivative?

Answer 13

because 2x+3 represents the slope at any point on f(x), this means that f ' (x) = 2x + 3

Answer 14

how do you find f(x)?

Answer 15

Take the integral?

Answer 16

yep or antiderivative

Answer 17

f(x) = x^2+3x+C

Answer 18

then you're told that the curve passes through (1,2)

so f(1) = 2

Answer 19

Uhm

2=1^2+3(1)+C?

C=-2?

Answer 20

Read this reply.....


I just wasted 2 seconds of your life.

Answer 21

Uhm thanks.

Answer 22

I guess.

Answer 23

C = -2 is correct

Answer 24

So y=x^2+3x-2?

Answer 25

yep

Answer 26

And I believe that 20 is E?

You plug in 3 to each and they both =5 which means the limit is the same as it approaches from the left and right, right?

Answer 27

careful

Answer 28

this is a potential trap

Answer 29

I recommend you derive each piece and examine the slopes of the tangent lines

Answer 30

............what

Answer 31

especially as x ---> 3

Answer 32

Why would I take the derivative of each?

Answer 33

the limit exists at x = 3, check
the function is continuous at x = 3, check
the function is differentiable at x = 3, ...hmm how do you check that?

Answer 34

let y = x+2

what is y' ?

Answer 35

Uhm I'm not sure

Also how do we know that I and II are true

Answer 36

y'=1

Answer 37

ok I guess I should start the problem over

Answer 38

Sure :P

Answer 39

plug x = 3 into each piece and you get y = 5 for each piece

that proves the limit at x = 3 exists

Answer 40

and because f(3) = 5, along with the fact that the limit is also 5, this proves that the function is continuous at x = 3

Answer 41

so statements I and II are true

Answer 42

But 3 isn't because the derivatives are different? So AKA the slopes are different?

Answer 43

exactly

Answer 44

the slope of the first piece is 1, while the second piece has a slope of 4

Answer 45

That abrupt change from 1 to 4 when you get to x = 3 (and pass it) proves it's NOT differentiable

Basically that sharp point "pops" the bubble of differentiability and makes it not the case. That's how a classmate described it one time and I like that analogy lol.

Answer 46

so it's actually D) I and II only

Answer 47

not differentiable at x = 3 specifically

Answer 48

Oh okay I understand

Answer 49

And 21 is a right? Because the graph crosses the x axis in those 2 places?

Answer 50

correct, you need to have a change in sign of f'' to represent a change in concavity for that point to be an inflection point

Answer 51

Wait what?

This is the graph of f''(x) and I know f''(x) shows concavity and points of inflection right?

Answer 52

yes, the given graph is f''

the roots of f'' are potential points of inflection

Answer 53

however, there needs to be a change from positive to negative (or vice versa) on f'' as it passes through the root for it to be an actual point of inflection

Answer 54

Okay thought so

Answer 55

I have no idea how to do 22 lol

Answer 56

that line is y = f ' (x)

Answer 57

we know 2 points that lie on that line (0,6) and (1,0)

Answer 58

so we can find the actual equation of f ' (x)

Answer 59

So that means f(x) is a parabola type graph

Answer 60

yes

Answer 61

And we know the parabola passes thorough (0,5) right?

Answer 62

Oh wait so we have 2 points and we can find the graph of f'(x) take the integral and find f(x) and plug in 1 right?

Answer 63

exactly

Answer 64

f'(x) = 6x+6

So that means f(x)= 3x^2+6x+C

Answer 65

But how do we find C

Answer 66

f(0) = 5

Answer 67

c=-105?

Answer 68

Dont think that's right lolol

Answer 69

seems way too small

Answer 70

c=5 :P

Answer 71

better

Answer 72

f(1)=3+6+5= 14 but thats not an answer

Answer 73

Oh wait sorry I put 3x^2 and not -3x^2

Answer 74

f(1)=8

Answer 75

yep d) 8

Answer 76

And 23 I have no idea how to do that... I think I kind of remember but not fully?

Answer 77

have a look at this page

http://www.sosmath.com/calculus/integ/integ03/integ03.html

Answer 78

Kind of rings a bell but I still cant remember how to do it

Answer 79

I know something that like... if the 0 and x^2 aren't in the right place you like make it negative times the integral or something?

Answer 80

On that page, you should see this

Answer 81

Okay?

Answer 82

Weird how I can't remember this because I think this topic was the one I got a 100 on the test on LOL

Answer 83

Let's define F(x) to be

\[\Large F(x) = \int_{0}^{x^2} g(t)\]

where

\[\Large g(t) = \sin(t^3)\]

Answer 84

Derive both sides of the first equation to get

\[\Large F^{\prime}(x) = \frac{d}{dx}\left(\int_{0}^{x^2} g(t)\right)\]

\[\Large F^{\prime}(x) = g(x^2)*\frac{d}{dx}\left(x^2\right)\]

\[\Large F^{\prime}(x) = \sin((x^2)^3)*2x\]

\[\Large F^{\prime}(x) = 2x\sin(x^{6})\]

Answer 85

again I'm using the rule found on that page I sent you

Answer 86

Yeah no I dont get it

Answer 87

Let me open my textbook and see what looks familiar

Answer 88

Oh right I see how to do the Fundamental Theorem I remember it

However the only problem is I'm not sure how to take the integral of sin(t^3) lol

Answer 89

Oh this is the 2nd fundamental theorem they want, let me look at that

Answer 90

The answer is C

Answer 91

close

Answer 92

you forgot about deriving x^2 to get 2x

Answer 93

2x sin x^6

Answer 94

yep

Answer 95

What do I do for 24?

Answer 96

how do we find slopes of tangent lines?

Answer 97

Derivative of f(x)