Question

Help!!!!!! please
The graph of the derivative f'(x) is given.
We know that f(0)=5
Find each of the following:
f(1),f(2),f(3),f(4),f(5),f(6),f(7)

Answer 1

@e.mccormick can you please help me with this problem

Answer 2

OK, can you find a function from the graph of the derivative, or is that where you ned the help?

Answer 3

I need help with all of it I need help finding all those f of's at the end of the problem. I dont know where to start

Answer 4

Notice how the derivative is composed of linear functions, i.e. straight lines. You can up with an equation for each and then anti-differentiate and use the given initial condition to find the value of C. That way you can find each of the values.

@farmergirl411

Answer 5

um still confused

Answer 6

@e.mccormick can you help me figure this out

Answer 7

are you still there

Answer 8

Yep. OK. The derivative is the slope of a line at a certain point. Right?

Answer 9

ok

Answer 10

You have this curve with very predicable slopes. They make a derivative that is straight lines. That is what genius12 was talking about. Each part of this derivative is a simple, linear equation.

Answer 11

So start by finding those equations.

Answer 12

Would the first equation be y=6x-1

Answer 13

Hmm, how did you get 6x-1?

Answer 14

Well, from 0 to -1 it si easy, that is \(y=-x\) In fact, the slope opn all of hese is easy. It is eiher -1 or 1.

Point slope form. \(y-y_1=m(x-x_1)\) and we know m.

Answer 15

well I was just looking at it closer should it be x+1 instead I don't know where I got that

Answer 16

For the first point, use (0,0) as the \(x_1,y_1)\) and you get y=-x. So lets use (1,-1) for the next point, with m=1

Answer 17

Do you see the second one, how this solves it?

Answer 18

y=1x-x is that the first equation

Answer 19

The first is just y=-x

\((y-0)=-1(x-0)\implies y=-x\)

Answer 20

oh ok I understand now I was just adding to many things in

Answer 21

The slope is easy in these because they are going up and down at the diagonals, so over one down one or over one up one, which makes the slope -1 for the parts where it goes down and 1 for the places it goes up. I sort of hinted at that, but thought I would state it.

Answer 22

For the second part, from 1 to 4, we have \(y-(-1))=-1(x-1)\) So what does that make the equation of the derivative there?

Answer 23

Second equation y=3x-2

Answer 24

Ummm....

Answer 25

Show us how you are getting these equations. The slope of a line in the form y=mx+b is rise over run... show us how you are getting 3x.

Answer 26

Where did that 3 come from?

Answer 27

are you still there

Answer 28

Yep. Site is being a bit laggy today....

Answer 29

@e.mccormick are you still there

Answer 30

I can tell it is laggy because you would not "at" mesage me if you had already seen me nswer you.

Answer 31

@farmergirl411 show us you are getting your equation y=3x-2

Answer 32

what answer nothing came through

Answer 33

oh ok I just refreshed my computer ot it now so then I am having a little trouble with equation two

Answer 34

or have you not said yet if equation 2 is right

Answer 35

y=3x-2 is not correct, but @farmergirl411 you need to show us how you are getting that equation.

Answer 36

\(y=3x-2\leftarrow \)that 3x. We wonder where the 3 popped up from. It is not right, but we want to find out where he mistake is being made so we can nip that in the bud and keep it from cropping up again!

Answer 37

because the line goes up three blocks

Answer 38

and across three blocks. Rise/run is the slope... rise is 3, run is 3.

Answer 39

right oh so then should the answer be 1 instead of 3

Answer 40

/cheer!

Answer 41

Algebra... the majority of mistakes I see and make myself all fall into basic algebra. The advanced stuff, more complex things, fine. But simple algebra will bite your anatomy on a regular basis. A close second is arithmetic mistakes.

Answer 42

y=x-2 then

Answer 43

OK. That is 2 down. Do you think you can do the third one?

Answer 44

ok hold on

Answer 45

Would it be the same equation y=x-2

Answer 46

No... you are looking at the top and guessing. Hehe. I can tell.

Answer 47

Use the point slope formula we just did.
\((y-y_1)=m(x-x_1)\) on the point \((4,2)\)

Answer 48

Well would it be y=x+2

Answer 49

\((y-2)=m(x-4)\)

Answer 50

y=4m+2

Answer 51

Now m in this case is \(-\frac{3}{3}=-1\)

Answer 52

y=-4m-2

Answer 53

http://assets.openstudy.com/updates/attachments/518056f7e4b0501a0d22b48a-farmergirl411-1367365395138-sw22.png

the rise of the third part is -3 as it drops down by 3 units. Run is again 3.

Answer 54

y=-4x-2

Answer 55

I don't know where you got a -4 from. Start again with it this way:
\((y-2)=-1(x-4)\)

Answer 56

ok y=-x-4

Answer 57

I am confussed where the 2 fits in

Answer 58

y=2x-4

Answer 59

Look at your graph. We are using the point (4,2) the 4 is the x coordinate, the 2 is the y.

Answer 60

\((y-2)=-1(x-4)\implies y-2=-x+4 \)
Now add 2 to both sides.

Answer 61

y=-x+2

Answer 62

y=-x+6

Answer 63

Yes. y=-x+6

Answer 64

sorry I keep adding wrong the last equation is what the 3 equation is

Answer 65

ok what do I do from this point to find all the f of's

Answer 66

OK. Now, there is one bit of good news. The derivative is nise and constant with no jumps, so we know this is a somewhat simple original equation. It is piecemeal (in parts) but that is about it.

Answer 67

So lets look at the parts of this one.

Answer 68

ok

Answer 69

so then what so I do next

Answer 70

\(y=-x\) for \(0\le x < 1\)
\(y=x-2\) for \(1\le x <4\)
\(y=-x+6\) for \(4\le x<7\)

Answer 71

Now, do you know how to do the anti-derivatives for each of those three equations?

Answer 72

no

Answer 73

Have you had anti-derivatives at all, or integration?

Answer 74

yes but I don't remeber how to do them

Answer 75

OK, do you remember deriviatives, like this one:
\[\frac{dy}{dx}x^n=nx^{n-1}\]

Answer 76

yes

Answer 77

but how is all of this getting me to find the list of f(1,2,3,4,5,6,7

Answer 78

Well, we need to do that in revese!
\[\int x^ndx=\frac{x^{n+1}}{n+1}\]

Once we have the calculations, the numbers will fall into place.

Answer 79

ok can you just work me through the problems cause I am starting not to understand and get lost

Answer 80

so then what does the n=

Answer 81

Yah, I thought you were sort of at the end of your rope there.

Answer 82

@e.mccormick can you help me I really need to get this problem done

Answer 83

are you still there

Answer 84

Wheeee.... site froze for a bit there.

Answer 85

oh ok I really need to get this problem done so can we hurry and get this problem done

Answer 86

Anything in calculus you don't remember, try reviewing these notes:
http://tutorial.math.lamar.edu/Classes/CalcI/CalcI.aspx

For a short, clear lecture on each topic, try these:
http://online.math.uh.edu/HoustonACT/videocalculus/

So we have these three formulas:
\(y=-x\)
\(y=x-2\) really means: \(y=x-2x^0\) for this part. The \(x^0\) will be important for integration.
\(y=-x+6\) really means: \(y=-x-6x^0\)

Now, I was talking about basic integration. Forgot one thing, the +C on the end.

\[\int x^ndx=\frac{x^{n+1}}{n+1}+C\]

Answer 87

ok

Answer 88

i think this problem is hard because you cannot really solve it without some additional assumption, namely that \(f\) is continuous

Answer 89

@satellite73 With that graph, it implies that it is.

Every \(x^1\) then becomes \(x^2\), but it needs to be multiplied by \(\frac{1}{2}\)

Every \(x^0\) becomes an \(x^1\) which is just x.

Answer 90

ok

Answer 91

So after I use that basic integration method, I get these:
\(y=-\frac{1}{2}x^2+C\)
\(y=\frac{1}{2}x^2-2x+C\)
\(y=-\frac{1}{2}x^2+6x+C\)

Answer 92

ok

Answer 93

We have to find what that C is, which is what satellite73 is talking about, and ths is where the question has a clue about it. f(0)=5 Well, f(0) means we use the first equation, so:

\(5=-\frac{1}{2}0^2+C\)

Well, 0 to anything times anything is 0. So:
\(5=C\)

Answer 94

OK. Ran out of stuff I typed up while the site was frozen. Hehe.

Answer 95

ok so then how do I find all of the f(1,2,3,4,5,6,)

Answer 96

Now I can replace those Cs and make this into a set of basic algebra problems.
If x is: \(0\le x < 1\) use:
\(y=-\frac{1}{2}x^2+5\)

If x is: \(1\le x <4\)
\(y=\frac{1}{2}x^2-2x+5\)

If x is: \(4\le x<7\)
\(y=-\frac{1}{2}x^2+6x+5\)