Question

Antiderivatives and sketching it.

Answer 1

Sketch the anti-derivative of f(x)=1/(1+x^4)
discuss maxima, minima, concavity and intervals where it is increasing or decreasing.

Answer 2

Firstly, there is no "the" antiderivative.
You can only find antiderivatives up to a constant.

Answer 3

Let \(\int f(x)\,dx = F(x)+C\)

\(F(x)\) is "an" antiderivative of \(f(x)\). By fundamental theorem of calculus, we have :
\[F'(x) = f(x)\]
Your goal is to sketch the graph of \(F(x)\),

Answer 4

Mhm.

Answer 5

So I don't have to find the antiderrivative for the one I am given?

Answer 6

that a very complicated integral to find. luckily your goal is not finding it
rather finding sketching and find extrema

Answer 7

Yeah, this problem is not about finding antiderivatives. This is just about sketching the graph of an antiderivative using the derivative

Answer 8

oh thank heavens. I had already begun trying to find it and I was dreading it.
okay, this just got much better :>

Answer 9

So what do I do?

Answer 10

as ganeshie said F'(x)=f(x)

Answer 11

F'(x) = 1/(1+x^4)

Notice that the derivative is always positive. What does this tell us about the graph of F(x) ?

Answer 12

f(x) is the derivative of F so look for the sign of f

Answer 13

you have learned that if f is an increasing function if f'>0

Answer 14

Isn't that
f(x) = 1/(1+x^4)
not f'(x) = 1/(1+x^4)
?

Answer 15

F'(x)=1/(1+x^4)
F is the anti-derivative
i just used f and f' to refresh your memory

Answer 16

but that's just symbols the meaning stays the same

Answer 17

i mean anti-derivative hehe made same mistake with "the"

Answer 18

xD aghh I am not understanding.

Answer 19

ok we have f>0 because 1/(1+x^4) >0 for any x
and f is the derivative of F
doesn't that mean that F is an increasing antiderivative

Answer 20

this is first derivative test if you remember

Answer 21

Right

Answer 22

we're looking to graph.. F(x)?

Answer 23

F is the function you usually start with and f is its own derivative
we use it to test for increasing, decreasing and critical points

Answer 24

yes!

Answer 25

in order to know the behavior of a function we use first derivative test and second derivative test

Answer 26

so to find the critical points, where F(x) crosses the x axis, we set f(x)=0?

Answer 27

usually that's what we do
but f cannot be zero because it is strictly positive

Answer 28

so here what do we do?

Answer 29

well what do you think?

Answer 30

definition of critical points points that make f'=0 or that make f' does not exist
in this case f' is f

Answer 31

so there are no critical points

Answer 32

Right o.o okay.

Answer 33

ok so we have F increasing
no critical points

Answer 34

not even at 0?

Answer 35

what is 0?

Answer 36

http://www.wolframalpha.com/input/?i=integral+of+1%2F%281%2Bx%5E4%29

Answer 37

that's inflation point

Answer 38

we were going to discuss that with second derivative test and concavity

Answer 39

oh ok, i see :)

Answer 40

so now you are ready to look for f'

Answer 41

which the second derivative

Answer 42

ooh. Man this upper case and lower case f are messing me up.

Answer 43

heheh those are just symbols
the problem is you focused on symbols like f' f
you should be flexible, pay attention to concepts

Answer 44

symbols just represent ideas, they are not something you should hold onto

Answer 45

siiigh haha yeah xD they're easy to get caught up on though.
ok, lemme find f'(x)

Answer 46

ok

Answer 47

- (4x^3)/(1+x^4)^2

Answer 48

yeah

Answer 49

set that to 0

Answer 50

for what x is f'=0

Answer 51

0

Answer 52

yes
and f'<0 for x?
f'>0 for x?

Answer 53

hm?

Answer 54

f'>0 for x<0 ===> concave up
f'<0 for x>0 ===> concave down

Answer 55

you can look at the sign of -4x^3

Answer 56

left of 0 this quantity -4x^3>0
right of 0 -4x^3<0

Answer 57

you got it

Answer 58

?

Answer 59

not really o.o

Answer 60

\[\sqrt[3]{-4}..?\]

Answer 61

but that's the same thing I get both both sides

Answer 62

no!
-4x^3=0 ===>x=0
where did you get radical -4

Answer 63

solving that for x?

Answer 64

but solving for x gives x=0

Answer 65

...so again, 0 is our point of inflection

Answer 66

yes

Answer 67

you got everything you need to graph F

Answer 68

so how do i find the specific points? and min and max?

Answer 69

I plug numbers into f'(x)?

Answer 70

there aren't any, we said there are not critical points

Answer 71

oh right right. Because there is no solution for f(x) to equal 0.

Answer 72

but to draw an accurate graph and see what points it goes through, I would plug numbers into f'(x)?

Answer 73

@inkyvoyd do you know how to do this? I want to draw a graph as accurately as possible. which function do i plug the point in to acheive that?

Answer 74

and also where it's increasing and decreasing.

Answer 75

@DannyO19

Answer 76

hey sorry, just saw this now xD

Answer 77

eem no I didn't try factoring anything.
I ended up just drawing a general sketch

Answer 78

oh nevermind, I haven't figured it out yet.. Do you have a specific question?

Answer 79

I guess I just wanted to get a graph as accurate as possible. but the question only calls for a sketch so i guess it's fine :)

Answer 80

What did you conclude from your sketch?

Answer 81

Did you just plug it in to your computer or did you actually use a formula to find the indefinite integral?

Answer 82

hm? well I don't have to find the intergral, because I only have to sketch it.

Answer 83

agh I have to sleep haha it's too late and i'm brain dead. But thanks so much :) maybe we can keep working on this tomorrow. I have early classes =.=

Answer 84

Of course...

Answer 85

It looks an integration of parts, which is good topic to review if you want to find the indefinite integral formulaically but since you can just graph it, it's simple right?

Answer 86

and also where it's increasing and decreasing.
we said it was increasing function all over R
we concludeed it has no extrema
we concluded it has an inflation point at (0,0)
we said it was concave up in the interval (-oo,0)
we saiid it was concave down in the interval (0,oo)
all the info you need to get the graph is here