prove lnx < square root of x if x > 0
ok you have to redefine them....$\displaystyle \ln(x)=\int_1^{x}\frac{1}{y}dy$ and $\displaystyle \sqrt{x}=\int_0^{x}\frac{1}{2y}dy$ $\displaystyle \forall{x}>0$ from there you can see what to do..but then again $\displaystyle \ln(x)\sim\sqrt{x}$...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
$\displaystyle \frac{1}{2\sqrt{x}} - \frac{1}{x} = 0$
$\displaystyle \frac{1}{2 x^{\frac{1}{2}}} = \frac{1}{x}$
$\displaystyle 2x^{\frac{1}{2}} = x$
$\displaystyle 2 = x^{\frac{1}{2}}$
$\displaystyle x = 4$
Use your first derivative test and you'll see it's a local/absolute minimum.
if $\displaystyle f(x)=\sqrt{x}-\ln(x)$ then $\displaystyle f'(x)=\frac{1}{2\sqrt{x}}-\frac{1}{x}$ 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$\displaystyle f''(x)=\frac{1}{x^2}-\frac{1}{4x^{\frac{3}{2}}}$ and we test the point 4 in there..since it gives$\displaystyle \frac{1}{32}$ which is positive $\displaystyle x=4$ is a minimum point on$\displaystyle (0,\infty)$
can you tell me if what I said was correct that since $\displaystyle \ln(x)\sim\sqrt{x}$ that you could make an argument that $\displaystyle \forall{x}>0,\ln(x)<\sqrt{x}$ is untrue?
Edit:forget it I dont know what I was thinking they are not asymptotically equivalent...forget I asked