Question

Hewp. Piece wise function as a "so far" function.

Answer 1

@ganeshie8 :)

Answer 2

this question is similar to a problem done a few days ago

http://openstudy.com/users/babynini#/updates/568f4480e4b0b283c160dfc3

Answer 3

ooh hmm..

Answer 4

How..do I go about doing this one?

Answer 5

this time i'm finding
f(t) = {
for each of the ones by f(x)?

if that makes sense

Answer 6

you are finding g(x) where

\[\Large g(x) = \int_{-4}^{x}f(t)dt\]

Answer 7

break it up into pieces

first find \[\Large \int_{-4}^{x}\sqrt{-t}dt\]

where x ranges from -4 to 0 (including -4 but not including 0)

Answer 8

so I find each of those points?

Answer 9

what is that integral equivalent to?

Answer 10

put x in place of t
integrating it= -(2/3)x^(3/2)

Answer 11

forget about the limits of integration for a moment

Answer 12

what is \[\Large \int \sqrt{-t}dt\] equal to?

Answer 13

(-t)^1/2.. ?

Answer 14

no

Answer 15

you are integrating (-t)^(1/2) to get something else

Answer 16

I'm not sure what answer you're looking for o.0

Answer 17

for now, I'm just looking for the result of \[\Large \int \sqrt{-t}dt\] or \[\Large \int (-t)^{1/2}dt\]

Answer 18

\[= -\frac{ 2 }{ 3 }(-t)^{3/2}\]

Answer 19

+C

correct

Answer 20

now evaluate it at the limits of integration, x and -4

what is \[\Large \int_{-4}^{x} \sqrt{-t}dt\] equal to?

Answer 21

\[-\frac{ 2 }{ 3 }(-x)^{3/2}+\frac{ 16 }{ 3 }\]

Answer 22

correct, so that takes care of the first piece of g(x)

Answer 23

what happens when x = 0 ?

Answer 24

what is the value of \[-\frac{ 2 }{ 3 }(-x)^{3/2}+\frac{ 16 }{ 3 }\] when x is zero?

Answer 25

It equals just 16/3

Answer 26

yep

Answer 27

so for just the first piece alone, the total area under the curve is 16/3

this will be used to form the next piece (because everything adds on cumulatively)

Answer 28

ahh so setting x = 0 is what will give us the actual area under the curve?

Answer 29

yeah because the first piece of f(x) spans from x = -4 to x = 0 (not including 0 itself)

Answer 30

so if \[\Large \Large g(x) = \int_{-4}^{x}f(t)dt\]

then g(x) is some piecewise function where the first piece is \[-\frac{ 2 }{ 3 }(-x)^{3/2}+\frac{ 16 }{ 3 }\] if -4 <= x < 0

Answer 31

the second piece of g(x) will be equal to \[\Large \frac{16}{3} + \int_{0}^{x}(t^2-4)dt\] where x is some value from 0 to 3 (not including 3 itself)

Answer 32

so in the second one I would set x equal to 0 again, and..also equal to 3?

Answer 33

first tell me what the antiderivative of `t^2 - 4` is

Answer 34

\[\frac{ t^3 }{ 3 }-4t\]

Answer 35

+ C

Answer 36

correct

Answer 37

now evaluate at the endpoints x and 0

Answer 38

\[\frac{ x^3 }{ 3 }-4x - 0\] like so?

Answer 39

looks good

Answer 40

so
\[\Large \frac{16}{3} + \int_{0}^{x}(t^2-4)dt\]
turns into
\[\Large \frac{16}{3} + \frac{x^3}{3}-4x\]
this is the second piece of g(x)

Answer 41

why don't we evaluate that second one at 3? even though it's not included there is still the space between say x being 2 and x being 3

Answer 42

we're getting there

Answer 43

we'll evaluate at x = 3 when we want to find the area from x = 0 to x = 3 to help set up the third and final piece

Answer 44

so what does \[\Large \frac{16}{3} + \frac{x^3}{3}-4x\] evaluate to when x = 3?

Answer 45

7/3

Answer 46

correct

Answer 47

this is the total area so far from x = -4 to x = 3

Answer 48

so the third piece of g(x) will look like

\[\Large \frac{7}{3} + \int_{3}^{x}(2t+4)dt \]

where x is now 3 or larger

Answer 49

t^2+4t +C

Answer 50

evaluate that at the endpoints (3 and x)

Answer 51

x^2+4x-21

Answer 52

good

Answer 53

so overall, this is g(x)

\[\Large g(x) = \int_{-4}^{x}f(t)dt\]

becomes

\[\Large g(x) = \begin{cases}
\Large -\frac{ 2 }{ 3 }(-x)^{3/2}+\frac{ 16 }{ 3 } \ \ \text{if } -4 \le x < 0\\
\Large \frac{16}{3} + \frac{x^3}{3}-4x\ \ \text{if } 0 \le x < 3\\
\Large \frac{7}{3}+x^2+4x-21\ \ \text{if } x \ge 3
\Large \end{cases}\]

Answer 54

after this point, you would graph g(x) as accurately as possible

Answer 55

so for that last point, to get a result without x's in it, would I evaluate at 3 again?
for the
7/3 + x^2+4x-21

Answer 56

Though I guess they're not really asking for the area under the curve, so I don't "technically" have to find it.

Answer 57

there is no upper bound though

Answer 58

g(x) is the area so far function

whatever x you plug in, it's the area from x = -4 to that inputted x value

Answer 59

Right, right, gotcha. It just goes off into infinity somewhere (or stops/meets again at some unknown point)

Answer 60

it will go off to infinity because `7/3 + x^2+4x-21` has the leading term of `x^2`

as x --> infinity, x^2 --> infinity

Answer 61

Yeah.
em, when trying to calculate points for this, I keep coming up with imaginary numbers..? haha

Answer 62

http://www.wolframalpha.com/input/?i=-%282%2F3%29%28-3%29%5E%283%2F2%29%2B16%2F3

Answer 63

keep in mind that the piecewise function is set up where the first piece is from -4 to 0 (3 is not in this interval)

Answer 64

right, but I put -3 in there. it does the same with -4 and -2

Answer 65

if you did x = -3, then the -x would be equal to +3

Answer 66

....riight xD

Answer 67

ok, well, I guess you can stick around to see the graph haha
but if not, thank you so much!! absolutely saved me lol

Answer 68

sure I can check that too. I graphed mine with geogebra if you wanted to see it

Answer 69

woah that would be awesome.

Answer 70

I'm drawing it by hand :| sigh

Answer 71

definitely better to use technology when you can

the first piece of g(x) is marked in blue, the second in red, and the third in green

Answer 72

all I did was graph the three functions (each piece as a full function) then restrict the domain of each function so that things line up

Answer 73

That's so cool. I haven't heard of that geogebra before.

Answer 74

Right, the homework calls for "draw an accurate sketch" so...idk if it's required to be by hand or not. It would be so much better with the tech. but idk o.o

Answer 75

you should check it out

http://www.geogebra.org/

Answer 76

it's hard to say for sure, but the teacher may accept hand drawn graphs

Answer 77

Thanks!
haha maybe I should turn in both xD

Answer 78

definitely not a bad idea

Answer 79

Yeah, i'll do that then. Thanks man :)

Answer 80

no problem