Math Question

[Voski]

The question is find the domain and then find the inverse of f(x) and then the domain of the inverse

so

f(x) = [3-e^(2x)]^(1/2) dom = (-inf, ln(3)/2)
f-1(x) = (1/2)ln(3-x^2)

for the inverse I think the domain should be (-sqrt(3), sqrt(3))

but the solution manual says [0, sqrt(3))

Any ideas or is it just an error on the manuals part?

[cotsi95]

a small phunnel is the best......



oh wait never mind

[indian_villager]

The question is find the domain and then find the inverse of f(x) and then the domain of the inverse

so

f(x) = [3-e^(2x)]^(1/2) dom = (-inf, ln(3)/2)
f-1(x) = (1/2)ln(3-x^2)

for the inverse I think the domain should be (-sqrt(3), sqrt(3))

but the solution manual says [0, sqrt(3))

Any ideas or is it just an error on the manuals part?

As far as the inverse is concerned the domain that an ln function can operate is [0,inf) Since you have 3-something in the ln function the domain of x can only go from 0-sqrt(3). Basically you can't have ln of a negative.

[Voski]

The question is find the domain and then find the inverse of f(x) and then the domain of the inverse

so

f(x) = [3-e^(2x)]^(1/2) dom = (-inf, ln(3)/2)
f-1(x) = (1/2)ln(3-x^2)

for the inverse I think the domain should be (-sqrt(3), sqrt(3))

but the solution manual says [0, sqrt(3))

Any ideas or is it just an error on the manuals part?

As far as the inverse is concerned the domain that an ln function can operate is [0,inf) Since you have 3-something in the ln function the domain of x can only go from 0-sqrt(3). Basically you can't have ln of a negative.

ya but it is -x^2 so
if x=-1 then
-(-1)^2 = -1
ln(3-1)=ln(2)

see what I am saying? The ln is positive from (-sqrt(3),sqrt(3)) atleast I think.

[infiniteslip]

It's funny tho cuz you can plot that f-1(x) and it exists from -sqrt(3) to sqrt(3) (endpoints not included). The x^2 term takes negative x>sqrt(3) back to positive and the log argument is still positive which is okay. But I think there's a catch in the x^2 term. Read http://en.wikipedia.org/wiki/Inverse_function towards the very end titles "Partial Inverses". They may be restricting the domain to x>=0.

[Bye bye now have fun]

when you invert it the range of f(x) becomes f(x)^-1's domain

[Voski]

when you invert it the range of f(x) becomes f(x)^-1's domain

So the issue is with the inverse function not being 1-1 once X<0?

[infiniteslip]

It's not 1 to 1 because of x^2. The inverse is supposed to take you backward, but with a square you don't know if you should go back to positive 1 or negative 1. There's two possibilities so you have to restrict the domain. It's one of those stupid little details that nobody remembers...

[Voski]

I didn't want to make a new thread so I am just gonna add to this one.

I have to take the integral of ln(2x+1)dx

here is what I did

a= 2x+1
1/2da=dx

u=ln(a)
du=(da)/a

dv=da
v=a

uv -integral (vdu)
= 1/2[aln(a)-integral(da)]

so I get
.5[(2x+1)ln(2x+1)] -x -.5 +c

but my question is does the (-.5) =c?

So the answer would be
.5[(2x+1)ln(2x+1)] -x +c

sorry if this makes no sense.

[Canon]

i miss high school (took college algebra there)

[K1024]

yeah the .5 would get absorbed into c. and i checked your work in maple and its all good. good jorb!

[Voski]

I think I am to tired to figure this one out I am just hitting a wall.

Here we go!

integral (arctan(4t)dt)

u=arctan(4t)
du = 4/(1+(4t)^2)dt

dv= dt
v= t

tarctan(4t) - integral (4t/(1+(4t)^2)dt)

I cant figure out how to take the integral of the second part. I'm gonna sleep maybe it will make sense in the morning. I need to buy a fucking solutions manual.

[K1024]

well ive completely forgotten how to do it...but i have maple so heres the answer!

t*arctan(4*t)-(1/8)*ln(1+16*t^2)

[Voski]

I woke up and it made sense in the morning. What is this maple thing?

[fineout]

voski...dont even ask..its an evil program that does nothing but ruin lives by doing math.

[K1024]

it does math better than any calculater you can buy, assuming your computer works better than said calculators.

similar to mathematica but better imo, it has its own language for programming and doing math in.

[Voski]

it does math better than any calculater you can buy, assuming your computer works better than said calculators.

similar to mathematica but better imo, it has its own language for programming and doing math in.

Does it show its "work"

[K1024]

if i remember right theres a way to get it to take steps...but those steps are often more convoluted than any human would think of doing...

its great for getting the answers though.

[fineout]

you have to know the steps pretty much, when i took calc2 i had to do things step by step in maple for projects, we had 10 of them with 10-15 problems in each and it ended up taking longer than just doing it myself. which is why i hate maple.

[Voski]

[attachment=3310:math_image.aspx.gif]

Can someone run this through maple?

I did it and got xln((x^2)+4) -2x +4arctan(x/2) +c

[Arcane]

a small phunnel is the best......



oh wait never mind

ahahahahaha!!!

[fineout]

arcane try this site, it does the integral things maple does

http://www.numberempire.com/integralcalculator.php Edited March 25, 2009 by fineout

[Reyomit]

I just ran it in mathematica and got the same thing, so unless you're looking for an answer with complex parts you're fine.

[K1024]

i can when i get home, but if it was ran through mathmatica with the same answer, then its probably right..

[K1024]

i did it and got the same answer.