Question

Concavity, extrema, asymptotes. Please check my work! :)

Answer 1

Question number 9

Answer 2

sorry about the first one being sideways!

Answer 3

@Kainui if you're free! :)

Answer 4

working on d and e still.

Answer 5

@amistre64 (:

Answer 6

Hmmm check your y-intercept again :)
I see a boo boo.

Answer 7

em em it's 0/sq-1

Answer 8

so it's - infinity?

Answer 9

Oh, there was a 0 on top ;O
Hmm that makes things tricky.

Answer 10

Have you looked at the graph? something like that xD

Answer 11

Ya, so no intercepts, right? :o
Doesn't cross x or y axis.
Hmm

Answer 12

OH I see why!

Answer 13

Yes, it WOULD cross at x=0,
But notice that x=0 is `not in the domain of this function`.

There is this "bad place" between 1 and -1.
Any value in there causes an imaginary number in the denominator.

Answer 14

ooo sneaky sneaky
so beneath those intercepts, I could write, f(t) would pass through these intercepts, however, since they are not in the domain of the function, we conclude that there is no x or y intercept" ??

Answer 15

For the x-intercept, I would keep this step,\[\large\rm 0=\frac{4t}{\sqrt{t^2-1}}\]And then add to it,\[\large\rm t\ne0,\text{ since 0 is not in the domain of t.}\]\(\large\rm \text{Therefore, no x-intercepts exist.}\)

Something like that maybe ^

Answer 16

The other one should be a little more straight forward.
Since t=0 is not allowed, we can't even plug it into the equation.
So no y-intercepts.

Answer 17

so just erase that entire y intercept part

Answer 18

Umm, ya.
Teacher maybe take a point for you plugging t=0 into the equation.
Not allowed to do that for this one.

Answer 19

okay! :)
"since t=0 is not allowed, there are no y-intercepts" yeah?

Answer 20

Maybe use similar language to what you used for the x-intercept :)
Just sounds nicer that way.

"Since 0 is not in the domain of f(t),
it follows that no y-intercepts will exist."

That's a correction you should make on the x-intercept as well,
the domain of `f(t)`, not t, my bad.

Answer 21

Either way sounds fine though ^^
So whichever wording you feel more comfortable with.

Answer 22

Critical points...
Oh did you negative exponent into product rule,
I see...
You're one of THOSE people lol

Answer 23

"those people"! excuse me, sir! xP
heh

Answer 24

Yes this language for the y-intercept is much better. The way I said it sounds like i'm in kindergarten haha

Answer 25

For some reason,
there are a bunch of people who HATE using quotient rule,
so they avoid it at all costs lol,
even when it's faster XD

I think it's fine either way though, in this case.

Answer 26

I love quotient rule, but product rule did seem simpler this time.

Answer 27

In part c, you say something about the denominator,
and then do some algebra it looks like, but with a small boo boo.\[\large\rm t^2-1=0\qquad\to\qquad t=\pm1\]Square root of a square gives us positive or negative 1.
Might be easier to understand if you just apply difference of squares formula,\[\large\rm t^2-1^2=0\qquad\to\qquad (t-1)(t+1)=0\]

Answer 28

except that 1 should not be squared *-*

Answer 29

1^2 = 1 silly :o

Answer 30

So I'm allowed to write 1^2 in place of 1

Answer 31

fiiine

Answer 32

but why would you want to? if
t^2-1=0 --> (t-1)(t+1)=0, so, t=-1,1
just the same?

Answer 33

Yes, that was just some justification for the plus/minus symbol magically showing up.

Answer 34

Ok the point I wanted to make though,
is that you should fix the domain, ya?

Answer 35

You do not ONLY have this restrition,\[\large\rm \sqrt{t^2-1}\ne0\]Since you have a square root,
you also have this restriction,\[\large\rm t^2-1>0\]The stuff under the root has to be positive.

Answer 36

Um um um ok ok let's get focused a sec...
Your derivative looks good.
It simplifies to this,\[\large\rm f'(t)=\frac{-4}{(t^2-1)^{3/2}}\]

Answer 37

Looking for critical points,\[\large\rm 0=\frac{-4}{(t^2-1)^{3/2}}\]
The denominator is giving us potential critical points of t=1, -1.
But since our domain is \(\rm (-\infty,-1)\cup(1,\infty)\)
we ignore them since these values are not in our domain.

Answer 38

Uh oh :O Brain eslope?
I can smell brains... oh boy.. lemme get the mop

Answer 39

ai ai ai

Answer 40

So i better write all that out =.= mew.

Answer 41

Maybe we should have focused on finding the domain at the very start,
because it's playing a major role in this problem D:

Answer 42

Yes indeed lol. Wait, I am writing this out cleverly and I shall show you a picture to see if that makes sense o.o

Answer 43

soowrie, super crappy shady picture haha