Calculus

[Shadows]

Anyone good at it?

I just need a lil help.

"Find the points of inflection and discuss the concavity of the graph of the function."

f(x) = 4/X^2+1

It seems easy but I always screw up somewhere.

-Thanks

 

[Bobby Sands]

 

[Flipmo]

 

[yak pac fatal]

LOL

you try googling the problem. i took pre calc in high school dont remember a damn thing.

 

[Shadows]

^yah, i tried google.

lol @ bobby posting Simon

 

[masta247]

does the "+1" part apply to the square/cube?

I mean is it 4/(X^2+1) or 4/X^(2+1) ?

Just draw that function and it'll be easy to read.

 

[Shadows]

i dont understand how they got the first

f'

than got to

f"

I'm not doing my derivatives correctly. :suspicious:

That's what I basically need help on.

 

[Sebastian]

Holy shit, math is horrible.

 

[masta247]

because:
f''(x) = 24/x^4

they just converted it because you can't divide by zero when x=0.

How did you get your f'(x)? Maybe it's the same thing, just written in a more complex way - at least if you use the correct formula.

 

[Da_Funk]

I'll help you when I get home from work tonight.

 

[Chronic]

The answer is: yes.

 

[2Pax]

Use the quotient rule to find f'(x) from f(x).

f(x) = g(x)/h(x)

f(x) = 4/(x^2+1) Hence g(x) = 4 and h(x) = (x^2+1)

Then f'(x) =

g'(x)h(x)-g(x)'h(x)
-----------------
h(x)^2

So f'(x) =

(0)(x^2+1)-(4)(2x)
-------------------
(x^2+1)^2


Hence =

-8x
-----------
(x^2+1)^2

Now do the same for f '' (x).

 

[Shadows]

EDIT: NVM. Thanks 2pax
I'll try that.

But i'll leave my mistake for u to see.



I guess i'm not just understanding the formula or doing a simple algorithmic error that i cant find.


F'(x) = (x^2+1)(1) - 4(2x)
I get: (x^2+1) - 8x
but the final is: -8x/(x^2+1)

Then that leads to:

F"(X) = (X^2+1)(-8X) - (8X)(X^+1)
(x^2+1)(-8) - (8x)(2)(x^2+1)(2x)

Why do I have to put (2) and (2x)?

Then i eventually get (which I know is wrong)
16x^3 + 24x^2 -8

when the answer was:
8(3x^2-1)/(x^2+1)^3

 

[2Pax]

F'(x) = (x^2+1)(1) - 4(2x)
I get: (x^2+1) - 8x
3

The '1' should be zero, the derivative of a constant is zero :cheesy:, I always make that mistake.

 

[2Pax]

Then that leads to:

F"(X) = (X^2+1)(-8X) - (8X)(X^+1)
(x^2+1)(-8) - (8x)(2)(x^2+1)(2x)

Why do I have to put (2) and (2x)?

Sorry about the shitty sizing.

 

[Shadows]

alright, I got the answer. Finally.

You a teacher or something?

 

[2Pax]

No lol, studying for a degree in electrical engineering and we covered quite a lot of calculus last year so it's still quite fresh in my mind.

 

[Cooper]

Wolfram Alpha is your friend.

http://www.wolframalpha.com/input/?i=4/(X^2%2B1)

 

[Cooper]

No lol, studying for a degree in electrical engineering and we covered quite a lot of calculus last year so it's still quite fresh in my mind.

Yea there's plenty of maths in taht degree :/

 

[2Pax]

Yea I knew there would be loads but even I didn't anticipate the amount involved. We're just starting The Fourier Series now.