Question

Graphs and integrals :)

Answer 1

(the first 4 graphs are choices for f)

Answer 2

I'm not sure how to go about doing the first part

Answer 3

for the first part, you integrate each piece as if it were a full function

if it helps, think of f(x) like this

f(x) = 0 if x < 0
OR
f(x) = x if 0 <= x <= 1
OR
f(x) = 2-x if 1 < x <= 2
OR
f(x) = 0 if x > 2

Answer 4

so g(x)=f'(x) ?

Answer 5

no

Answer 6

no derivatives are involved here

Answer 7

hrrm

Answer 8

can you show me what you mean? xD
like with the
0</=x</=1 one

Answer 9

Let's focus on the piece where \(\Large 0 \le x \le 1\)

So that means f(x) = x and f(t) = t

so

\[\Large g(x) = \int_{0}^{x}f(t)dt\]

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

\[\Large g(x) = \frac{1}{2}t^2+C{\Huge]}_{0}^{x}\]

\[\Large g(x) = \left(\frac{1}{2}(x)^2+C\right)-\left(\frac{1}{2}(0)^2+C\right)\]

\[\Large g(x) = \frac{1}{2}x^2\]


keep in mind that this is just for the piece where \(\Large 0 \le x \le 1\)

Answer 10

you'll do the same for the other pieces

Answer 11

(sorry, open study kicked me off for a bit haha.)

Answer 12

that's fine

Answer 13

hm, ok. I think I see o.o

Answer 14

how do the numbers 0 and 1 affect it then?
0 < x <1

Answer 15

they just are restrictions on the domain. They don't affect the integral

Answer 16

imagine you programmed the calculator to only show the region from x = 0 to x = 1. As far as the calculator knows, f(x) is equal to x. Anywhere else this isn't true, but for this small window, it is true.

Answer 17

Ahh gothca. I see.
but then later when the restriction changes to 1<x</=2 it does affect it?

Answer 18

then the function f(x) changes to 2-x

so you'll have f(t) = 2-t

integrate with respect to t and use the endpoints 0 and x to get g(x) for that specific interval

Answer 19

2-[(1/2)x^2] ?

Answer 20

should be 2x - (1/2)x^2

Answer 21

whoops, typo xD
and then the first one is the same as f(x) right? it is 0
and the last one is 1

Answer 22

yeah the integral of 0 is 0

Answer 23

the last one is also zero for the same reason

Answer 24

the last one is 0? hm ok

Answer 25

yeah it's the same as the first box

Answer 26

another quick question, about f(x)
I know which graph it is, but where is it differential?

Answer 27

*differentiable

Answer 28

I would expect that it is everywhere except for the peak

Answer 29

(-infinity, 1)u(1,infinity) but that is incorrect

Answer 30

you're asking where f(x) is differentiable?

Answer 31

Yeah

Answer 32

well each piece is a polynomial, so each piece is differentiable if you focus on one piece at a time. However, if you look at the junction between each piece, then it's a different story

Answer 33

what I would do is derive f(x), ie derive each piece, and look to see if f ' (x) is continuous at the junction points

Answer 34

do you think I could just put (-infinity, infinity)? if each point itself is differentiable

Answer 35

nope that didn't work haha neevermind

Answer 36

what did you get for f ' (x) ?

Answer 37

from the graph?

Answer 38

it should look like.. two horizontal lines, right?

Answer 39

no based on the piecewise definition

Answer 40

I'm not sure how to do it from there?

Answer 41

oh true, yes the graph of f ' (x) will be 2 horizontal lines. Well 3 actually but the x axis covers up that third line

Answer 42

so that shows you where f(x) isn't differentiable. Where you have those jump discontinuities on f ' (x)

Answer 43

which is at 1 and..

Answer 44

derive 2-x

Answer 45

2x-(1/2)x^2 ?

Answer 46

(that didn't work as an answer for g(x) btw o.o)

Answer 47

oh right, I'm not thinking

Answer 48

Let's first graph f(x)

blank xy axis

Answer 49

first piece

f(x) = 0 when x < 0



shown by the darker line

Answer 50

next piece is f(x) = x when 0 <= x <= 1