Question

plz help me with this one, plz provide the work or start with taking the derivative/double derivative of this.
y=x^2 +1/x^2 -1
find the relative extrema, points of inflection, intervals where increasing/decreasing, intervals where concave up/down and any vertical or horizontal asym. show all work and sign charts, and tests used.

Answer 1

@DanJS

Answer 2

Hey Katie :)
The problem looks like this?
\[\large\rm y=\frac{x^2+1}{x^2-1}\]Quotient rule would be a good place to start, ya?

Answer 3

hi yes, looks just like that and yeah use quotient rule for that and take the first derivitive to start with

Answer 4

I like to "set up" the rule before applying it.
Nice way to stay organized, at least early on in calculus.\[\large\rm y'=\frac{\color{royalblue}{(x^2+1)'}(x^2-1)-(x^2+1)\color{royalblue}{(x^2-1)'}}{(x^2-1)}\]So the blue stuff is what we need to differentiate, ya?
Simple power rule gives us:\[\large\rm y'=\frac{\color{orangered}{(2x)}(x^2-1)-(x^2+1)\color{orangered}{(2x)}}{(x^2-1)}\]

Answer 5

yes, i got that part. neat work i must say!

Answer 6

Woops! Denominator should've been squared :) silly mistake on my part.

Answer 7

yes (x^2 -1)^2 sorry i should have pointed it out

Answer 8

For extrema, we set our first derivative to zero,\[\large\rm 0=\frac{(2x)(x^2-1)-(x^2+1)(2x)}{(x^2-1)^2}\]
The numerator is going to give us our traditional `critical points`,
locations of horizontal tangency,

while the denominator will tell us where the function has vertical tangency.
\(\large\rm 0=(x^2-1)^2\qquad\to\qquad x=\pm1\)
Notice that x=1 and x=-1 are not in the `domain` of this function, so we ignore those values here.

Answer 9

Hmm they wanted us to talk about asymptotes?
Maybe we should have done that first, before taking derivatives.
Either way is fine I suppose :d

Answer 10

When we set our numerator equal to zero,\[\large\rm 0=(2x)(x^2-1)-(x^2+1)(2x)\]What critical points do you get? :)

Answer 11

okay wait let me tell u the HA and VA that i got

Answer 12

Oh sure.

Answer 13

y=1 HA?

Answer 14

\[\large\rm \lim_{x\to\infty}y=1\]Ok that sounds right so far!
As we go far far to the right (or left), our function is approaching this line y=1.

Answer 15

okay and VA in this case would be 0+1/0-1 so 0?

Answer 16

wait sorry that's the y-inter

Answer 17

okay VA= 1 and -1

Answer 18

Denominator can never be zero,\[\large\rm x^2-1\ne0\qquad\implies\qquad (x-1)(x+1)\ne0\]\[\large\rm x\ne1,\qquad x\ne-1\]Ok great.

Answer 19

okay lets cont with the direvitives! tysm so far

Answer 20

You have your first derivative, setting it equal to zero so we can find critical points,\[\large\rm 0=\frac{(2x)(x^2-1)-(x^2+1)(2x)}{(x^2-1)^2}\]Multiply both sides by (x^2-1)^2 to get rid of the denominator,\[\large\rm 0=(2x)(x^2-1)-(x^2+1)(2x)\]Now solve for x,
what do you get for critical points? :)

Answer 21

(2x^3 -2x)- (2x^3+2x)

Answer 22

Hmmm, you can do that I suppose.
But I would instead recommend factoring, unless that's going to be too confusing.

Answer 23

\[\large\rm 0=\color{royalblue}{(2x)}(x^2-1)-(x^2+1)\color{royalblue}{(2x)}\]\[\large\rm 0=\color{royalblue}{(2x)}\left[(x^2-1)-(x^2+1)\right]\]

Answer 24

And then apply your `Zero-Factor Property`:\[\large\rm 0=2x,\qquad\qquad 0=(x^2-1)-(x^2+1)\]

Answer 25

x=0 and x=1 and x=-1

Answer 26

x=0 sounds good!
1 and -1, hmmm let's see...
distributing that negative sign gives us:\[\large\rm 0=x^2-1-x^2-1\]The x^2's subtract out,\[\large\rm 0=-2\]

Answer 27

So it looks like we're not getting any critical points from this factor, ya? :o

Answer 28

wait i am reading over your stuff

Answer 29

no, the ones factored are our zeros right

Answer 30

cause numberator are the critical zeros right

Answer 31

Zero-Factor Property tells us when we have things multiplied together giving us zero, we can set each individual factor equal to zero:\[\large\rm 0=\color{royalblue}{(2x)}\color{orangered}{\left[(x^2-1)-(x^2+1)\right]}\]
\[\large\rm \color{royalblue}{0=2x},\qquad\qquad \color{orangered}{0=\left[(x^2-1)-(x^2+1)\right]}\]This blue piece gave you x=0, good.
I'm trying to say that this orange piece will not give you x=1 and -1.
Hopefully the algebra I did before was clear, I skipped a couple steps maybe.

Answer 32

The orange is actually giving us `no critical points`.
We only have the critical point at x=0.

Answer 33

Let's see if we can figure out increasing/decreasing :)

Answer 34

brb one sec, i need some peanut butter :O

Answer 35

right i see what u are saying. x^2 -1 -x^2-1=0 the X^2 got cancled but the -1 and -1 becomes -2 so because the x are alredy cancled right?

Answer 36

yeah go ahead i am not leaving untill i get this solved. haha

Answer 37

tysm for helping me! :)

Answer 38

If you're more familiar with doing a sign chart, we can try that.
I don't quite remember how to do that though... I always did the interval method.
They're essentially the same though, I think.

Answer 39

yes sign charts plz i am familiar

Answer 40

\[\large\rm y'=\frac{(2x)(x^2-1)-(x^2+1)(2x)}{(x^2-1)^2}\]Notice that our denominator is being squared.
It will ALWAYS be positive.
That will have no effect on the `sign` of our derivative, so we can ignore it.\[\large\rm y'=(2x)(x^2-1)-(x^2+1)(2x)\]This is kind of sloppy notation saying that y' equals this ^
But whatever, it simplifies things for us.

Answer 41

We need to know what is happening on the left and right side of .... zero,
since that is our critical point.

Answer 42

right but did u read my commet about -2 is that right the way i explained that x^2 gets cancled that's why we do not get a critical point there?

Answer 43

Hey @blueskies :) Don't hijack someone else's question.
I see you have your question open. In the chat box below your question type @zepdrix or someone else's name like @Jhannybean and it will alert them that you need assistance.
I'll try to take a look if I get a chance.

Answer 44

thank you @zepdrix :)

Answer 45

Ya that's a good way to explain it Katie.
You actually end up with 0=-2 after you cancel out the x^2's.
And this is a FALSE statement.
0 can not equal -2, so this is giving us no information, we can't have a zero here.

Answer 46

right so that concludes 0 being the only critical point. okay you can carry the sign chart now. sorry to ask that agian.