proof problem( I dont know what this belong to)

**soleilion** · April 5th 2008, 12:53 PM

prove lnx < square root of x if x > 0

**Mathstud28** · April 5th 2008, 01:07 PM

ok you have to redefine them.... $\ln(x)=\int_1^{x}\frac{1}{y}dy$ and $\sqrt{x}=\int_0^{x}\frac{1}{2y}dy$ $\forall{x}>0$ from there you can see what to do..but then again $\ln(x)\sim\sqrt{x}$ ...so you could make a bid that this is untrue

**soleilion** · April 5th 2008, 01:20 PM

Originally Posted by Mathstud28

ok you have to redefine them.... $\ln(x)=\int_1^{x}\frac{1}{y}dy$ and $\sqrt{x}=\int_0^{x}\frac{1}{2y}dy$ for every $x>0$ from there you can see what to do..but then again $\ln(x)$ ~ $\sqrt{x}$ ...so you could make a bid that this is untrue

sorry, i misunderstood
can u see this?
http://img99.imageshack.us/img99/5271/13493377zy4.png

this is how to prove by solution manual
I dont understood how the minimum is 4

can u tell me the reason?
thank you

**o_O** · April 5th 2008, 01:28 PM

$\frac{1}{2\sqrt{x}} - \frac{1}{x} = 0$
$\frac{1}{2 x^{\frac{1}{2}}} = \frac{1}{x}$
$2x^{\frac{1}{2}} = x$
$2 = x^{\frac{1}{2}}$
$x = 4$

Use your first derivative test and you'll see it's a local/absolute minimum.

**Mathstud28** · April 5th 2008, 01:30 PM

if $f(x)=\sqrt{x}-\ln(x)$ then $f'(x)=\frac{1}{2\sqrt{x}}-\frac{1}{x}$ find the places where it is undefined and zero..this is at x=0(undefined) and x=4(zero) since 0 is not in the domain we can get rid of it..now we need to check where the original function is undefined...it is undefined nowhere on the domain of the question...so now all we have is the point x=4...we test that in the second derivative $f''(x)=\frac{1}{x^2}-\frac{1}{4x^{\frac{3}{2}}}$ and we test the point 4 in there..since it gives $\frac{1}{32}$ which is positive $x=4$ is a minimum point on $(0,\infty)$

**Mathstud28** · April 5th 2008, 06:59 PM

can you tell me if what I said was correct that since $\ln(x)\sim\sqrt{x}$ that you could make an argument that $\forall{x}>0,\ln(x)<\sqrt{x}$ is untrue?

Edit:forget it I dont know what I was thinking they are not asymptotically equivalent...forget I asked

**Krizalid** · April 6th 2008, 05:55 AM

Originally Posted by Mathstud28

$\sqrt{x}=\int_0^{x}\frac{1}{2y}dy$

Careful when workin' with such parameters, this one is wrong.

**Moo** · April 6th 2008, 05:59 AM

Hi,

Integrate from 1 to x, uh ?

**Mathstud28** · April 6th 2008, 07:32 AM

Integrate from 1 to x for what? and Krizalid is that wrong because the boundary restriction of that does not mesh well with this problem?

**Moo** · April 6th 2008, 07:37 AM

..

$\int_0^x \frac{1}{2y} dy = \frac{1}{2} \ln(x) \big|^x_0$

Which is not defined at 0.

Plus $\ln(\sqrt{x})$ is not equal to $\sqrt{x}$

**Mathstud28** · April 6th 2008, 07:40 AM

I am sorry that is a typo it should read $\sqrt{x}=\int_0^{x}\frac{1}{2\sqrt{y}}dy$ ...sorry for the typo

**Moo** · April 6th 2008, 07:48 AM

I think that you should sometimes leave integrals to solve problems the way it's good to solve

**Mathstud28** · April 6th 2008, 07:50 AM

I am not quite sure what you mean by that..but ok! haha =D

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