Sketch the graph of the given function , :) help?

[Queenisabella87]

Hey guys,

Given the function
f(x) =

1-x[sup:2fil1e88]2[/sup:2fil1e88]
______
x[sup:2fil1e88]2[/sup:2fil1e88]+1

Sketch the graph,

Make sure to label the zeros, critical points and inflection points as well as any asymptotes (if they exist).



Okay so what I did here is

I found the derivative (quotient rule) and set it equal to 0. but I think I got the wrong critical points.

can anyone guide me with their work so I can check it against mine?

 

[Deleted member 4993]

Hey guys,

Given the function
f(x) =

1-x[sup:26aueal2]2[/sup:26aueal2]
______
x[sup:26aueal2]2[/sup:26aueal2]+1

Sketch the graph,

Make sure to label the zeros, critical points and inflection points as well as any asymptotes (if they exist).



Okay so what I did here is

I found the derivative (quotient rule) and set it equal to 0. but I think I got the wrong critical points. <<< What did you get?

can anyone guide me with their work so I can check it against mine?

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[Queenisabella87]

I can't get past finding the derivative and setting equal to Zero to find critical points...

Pretty harsh of you don't you think? I'm here for help, I don't know how to start this specific problem.

 

[lookagain]

I can't get past finding the derivative and setting equal to Zero to find critical points...

Given the function

\(\displaystyle f(x) \ = \ \frac{1 - x^2}{x^2 + 1}\)


Sketch the graph.

Make sure to label the zeros, critical points, inflection points, as well as any asymptotes (if they exist).


Okay so what I did here is

I found the derivative (quotient rule) and set it equal to 0. but I think I got the wrong critical points.

Can anyone guide me with their work so I can check it against \(\displaystyle mine \ ( your \ alleged \ work)?\)

Queenisabella87,

here is an analogy:

We want your work shown (that you allege to have) in exchange for our assistance to you.

As it is now, you want us to give our assistance to you based in part of your "promise"
(claim of having some work done for the problem). We would need your actual work
(so far) "delivered to us" in the forum first, before we "hand over" our assistance.

 

[Deleted member 4993]

I can't get past finding the derivative and setting equal to Zero to find critical points...

Pretty harsh of you don't you think? I'm here for help, I don't know how to start this specific problem.

What is so harsh about asking you to show your work?

You say "I don't know how to start this specific problem" - and then you say you "found derivatives and critical points".

Do you see why we don't know where to start to help you?

You have been in this board for a while - you know the rules.

 

[galactus]

If you found the derivative, then you should have gotten

\(\displaystyle f'(x)=\frac{-4x}{(x^{2}+1)^{2}}\)

There is only one value that makes this equal to 0.

What x value makes -4x=0?.

You can see it on the graph.

To find any inflection points, find the second derivative, set the numerator equal to 0 and solve for x.

There are two inflection points. You can see whereabouts they are on the graph.

To find the horizontal asymptote, think about what f(x) approaches as x gets bigger and bigger.

When the power of the numerator is equal to the power of the denominator, as in this case, you can find the HA by

noting \(\displaystyle \frac{ax^{n}}{bx^{n}}=\frac{a}{b}\)

 

[Queenisabella87]

Thank you Galactus<3 I got it!!

You're my hero!!!!!!!!! :!: