calculus help

[pwnt by pat 10+ year member]

So I'm a little stuck on my calculus work. I just want to see if anyone can verify what I found for my results.

We're doing inverse functions:

f(x)= (4x-1) / (2x + 3)

f'(x)= [4(2x +3) - 2(4x-1)] / (2x+3)^2 = 14/(2x+3)^2

and f-1(a)= 1/f'(f-1(a))

Actually, while typing this out, I found my answer doesn't mean shit....

Can anyone give me a hand? My textbook and class notes are far less than useful.

Click to read more...

 

[jeffs42885 10+ year member]

what college do you goto, that loks kinda familiar //content.invisioncic.com/y282845/emoticons/biggrin.gif.d71a5d36fcbab170f2364c9f2e3946cb.gif

 

[pwnt by pat 10+ year member]

Oh, I'm sure you know....

 

[lilmaniac2 5,000+ posts]

i hated calc

 

[LinxuS 10+ year member]

So I'm a little stuck on my calculus work. I just want to see if anyone can verify what I found for my results.
We're doing inverse functions:

f(x)= (4x-1) / (2x + 3)

f'(x)= [4(2x +3) - 2(4x-1)] / (2x+3)^2 = 10/(2x+3)^2

and f-1(a)= 1/f'(f-1(a))

Actually, while typing this out, I found my answer doesn't mean shit....

Can anyone give me a hand? My textbook and class notes are far less than useful.

First of all..

4(2x +3) - 2(4x-1) = 14

= 14/(2x+3)^2

don't forget to distribute the negative sign.

I'm not really understanding the notation though for inverse functions?

 

[pwnt by pat 10+ year member]

I like it, just not inverses. Never got it in pre-algebra in highschool either.

 

[pwnt by pat 10+ year member]

edited.

notation is

f-inverse of (a) = 1/f'(f-inverse(a)) er... That's what my lecture notes say at least.

g'(a)= 1/f'(f-inverse(a)) is what the book says.

g(a)= f-inverse(a), so g'(a) would be the derivative of f-inverse(a)

 

[LinxuS 10+ year member]

Had a second edit, it = 14 btw.

We barely went through inverse functions since theyre not on the AP exam, or theres only like 1 question.

I'm stuck here too, and now it's pissing me off lol.

 

[pwnt by pat 10+ year member]

I understand finding g'(a) when a is given.... but I don't know

 

[dinty01 10+ year member]

i used to be into doin calc for fun, yea I know thats sick but I enjoyed it for some reason, now im into computers, yea binary number system and hexadecimal number system, yea . . . . anyways wish I could remember enough of it to help ya out, i got lost here: f'(x)= [4(2x +3) - 2(4x-1)] / (2x+3)^2 = 10/(2x+3)^2, o well i guess the old saying is true, if ya dont use it ya lose it, good luck with it

 

[pwnt by pat 10+ year member]

hex sucks.

That's just the difference forumla for differentiation

 

[pwnt by pat 10+ year member]

Well, I got to:

g(a)= .5*sqrt(1/g'(a)) - 3....

 

[dleccord 5,000+ posts]

what, are you guys embarass to run into each?