Question

find the value of x>1 for which the function F(x)=integral x to x^2 [(1/t)*{ln([1-t]/32}]dt is increasing and decreasing

Answer 1

find the value of x>1 for which the function \[F(x)= \int\limits_{x}^{x^2}(1/t)*\ln [(1-t)/32]dt \] is increasing and decreasing

Answer 2

F'(x) = 2x [ (1/x^2) ln(1-x^2)/32 ]

find where f'(x) is positive or negative

Answer 3

@sourwing i think there should be an \(x\) in the denominator, not that it makes much difference

Answer 4

ah yes, I was just noticed the x lol

Answer 5

i am also getting a slightly different derivative, but my algebra could be off

Answer 6

let me do the derivative again

Answer 7

i get as a first step
\[\frac{\log(1-x^2)}{16x}-\frac{\log(1-x)}{32x}\]

Answer 8

as a second step
\[\log((1-x^2)^2)-\log(1-x)\] ignoring the denominator which is positive in any case

Answer 9

but like i said, my algebra could be not so good at this hour

Answer 10

F(x) = ∫{x->2} (1/t) ln(1-t^2)/32 + ∫[2->x^2] (1/t) ln(1-t^2)/32
F(x) = -∫{2->x} (1/t) ln(1-t^2)/32 + ∫[2->x^2] (1/t) ln(1-t^2)/32
F'(x) = - (1/x) ln(1-x)/32 + (1/x^2) ln(1-x^2)/32 * 2x

Answer 11

no matter, i think we lost the person answering the question

Answer 12

yeah that is what i got too
then cancelled an \(x\) in the second term, added up the fractions and got the numerator i wrote above

Answer 13

@satellite73 is it not that (integral (f (x)))' = f(x)?

Answer 14

@loser you have an \(x\) at both ends of the limit of integration

Answer 15

\[\int _a^{x^2}f(t)dt)'=f(x^2)\times 2x\] by the chain rule

Answer 16

F'(x) = ln(1-x)/(32x)

Answer 17

oh wait no

Answer 18

i get this as a first step
\[\frac{\log(1-x^2)}{16x}-\frac{\log(1-x)}{32x}\]

Answer 19

by \(\frac{2x}{32x^2}=\frac{1}{16x}\)

Answer 20

yep got the samething

Answer 21

@Loser66 the derivative of
\[\int_a^xf(t)dt\] is \(f(x)\) but via the chain rule the derivative of
\[\int_a^{g(x)}f(t)dt\] is
\[f(g(x))g'(x)\]

Answer 22

then adding up the fractions gave me \[\log((1-x^2)^2)-\log(1-x)\] as a numerator

Answer 23

property of logs gives
\[\log((1+x)^2(1-x))\]

Answer 24

yw
@Loser66 also you have an \(x\) as the lower limit of integration, so you have to subtract off \(f(x)\)

Answer 25

@satellite73 shoult it be ln[ (1-x^2)^2 / (1-x) ]?

Answer 26

it should be but
\[\frac{1-x^2}{1-x}=1+x\] so i cancelled one of them

Answer 27

ahh ..

Answer 28

not really sure it makes it easier, but i figured it probably would
question is still for which \(x\) is that thing positive

Answer 29

i guess all the work is solving
\[(1+x)^2(1-x)>1\] for \(x\) but maybe there is a better method

Answer 30

oh no, it works out well
get \[-x^3-x^2+x > 0\]

Answer 31

end up solving a quadratic equation
all this work and the person is gone
too bad

Answer 32

lol

Answer 33

don't we need to take the domain into account?

Answer 34

the denominator is unimportant since \(x>1\) is given

Answer 35

i think you end up with the golden mean for this one

Answer 36

\[x (x^2+x-1)<0\] is a first step

Answer 37

well, I thought F(x) is only differentiable on (-1,0) U (0,1)?

Answer 38

not sure why
problem states \(x>1\) so in integrand is defined for all \(x\) if \(x>1\)

Answer 39

I'm pretty sure there is a typo.

Answer 40

well, i am not going to fret about it since i am just fooling around
but i did think it was rather cool that you end up with the golden mean
\[1<x<\frac{-1+\sqrt5}{2}\]

Answer 41

urg... screw it :/ i'm off playing game lol

Answer 42

where did u got
\[[\ln (1-x ^{2})/16x]-[\ln (1-x)/32x]\] from pl explain

Answer 43

pl help me how u found f'(x)