i made a limit on both infinity and minus infinity for them

and i tried to find but its not working

http://img201.imageshack.us/img201/5458/23597303em5.gif

Printable View

- Dec 20th 2008, 08:26 AMtransgalactichow to find the max,min,sup,inf of these cases..
i made a limit on both infinity and minus infinity for them

and i tried to find but its not working

http://img201.imageshack.us/img201/5458/23597303em5.gif - Dec 20th 2008, 09:55 AMbkarpuz
The domain of the function $\displaystyle \mathrm{ln}(x)$ is $\displaystyle (0,\infty)$, with $\displaystyle \lim_{x\to0}\mathrm{ln}(x)=-\infty$.

Let

$\displaystyle f(x):=\frac{1}{1+(\mathrm{ln}(x))^{2}},$

then

$\displaystyle f^{\prime}(x):=-\frac{2\mathrm{ln}(x)}{x(1+(\mathrm{ln}(x))^{2})^{ 2}},$

which implies $\displaystyle f^{\prime}(1)=0$.

Therefore its maximum value is $\displaystyle f(1)=1$.

On the other hand, $\displaystyle f$ attains its inferior value as $\displaystyle x\to\infty$ (as you have shown).

For the second one, let

$\displaystyle g(x):=x+\frac{1}{x}$

for $\displaystyle x>0$ (I believe it must be given on the positive halfline).

Clearly, as $\displaystyle x\to0$ or $\displaystyle x\to\infty$, we have $\displaystyle g(x)\to\infty$ (superior value).

And on the other hand, we have

$\displaystyle g^{\prime}(x)=1-\frac{1}{x^{2}},$

which implies $\displaystyle g^{\prime}(1)=0$

and thus $\displaystyle g(x)\geq g(1)=2$ for all $\displaystyle x>0$ (minimum value).

You may similarly show the third one too, I have to say that this one is very interesting, just show that its maximum values decrease and similarly show that its minimum values increase. - Dec 20th 2008, 11:24 PMtransgalactic
so the max and the sup of the first one is 1

what is the min and inf of the first one? - Dec 21st 2008, 12:41 AMtransgalactic
$\displaystyle

g(x):=x+\frac{1}{x}

$

the maximal number is infinty

when x->+infinty

the smallest value is found when

x->-infinity

?? - Dec 21st 2008, 01:09 AMtransgalactic
regarding the third one:

the derivative is

(cosx * x - sin x)/x^2

how to find the extreme points in order to find max min - Dec 21st 2008, 01:42 AMbkarpuz
$\displaystyle \lim_{x\to\infty}\ln(x)=\infty$ implies $\displaystyle \lim_{t\to\infty}\frac{1}{1+(\ln(x))^{2}}=0$ which is the

**inferior**value, because there is no $\displaystyle x_{0}$ point that $\displaystyle \frac{1}{1+(\ln(x_{0}))^{2}}=0$ holds.

As I said before $\displaystyle x+\frac{1}{x}$ is unbounded; i.e., its**superior**value is $\displaystyle \infty$.

And it takes its**minimum**value at $\displaystyle x=1$ with $\displaystyle 1+\frac{1}{1}=2$.

**Note**. To find the extreme points of a function $\displaystyle f$, just find the zeros of the equation $\displaystyle f^{\prime}=0$.

Say $\displaystyle x_{0}$ is such a point that $\displaystyle f^{\prime}(x_{0})=0$, then if $\displaystyle f^{\prime\prime}>0$ we say that $\displaystyle f$ has a minimum point at $\displaystyle x_{0}$, and if $\displaystyle f^{\prime\prime}<0$ we say that $\displaystyle f$ has a maximum value at $\displaystyle x_{0}$.

**PS**. I just now saw that it seems hard by proving in the way I told you before. - Dec 21st 2008, 02:08 AMtransgalactic
in the third case the sinus function

i get

cosx * x -sinx =0

how to solve this equation?? - Dec 21st 2008, 02:31 AMmr fantastic
An obvious solution is x = 0. The function $\displaystyle f(x) = \frac{\sin x}{x}$ has a removable singularity at x = 0. It makes sense to define f(0) = 1.

Then the global maximum of $\displaystyle f(x) = \frac{\sin x}{x}$ is located at (0, 1).

The other local maxima (and local minima) cannot be found algebraically. You will need to use a numerical procedure to get approximate solutions. - Dec 21st 2008, 02:38 AMbkarpuz
Yes it is not easy, thus, we have to find other ways to find max and min values of this function.

May the the following analysis may be helpful to you.

Let $\displaystyle f(t):=\sin(t)/t$ for $\displaystyle t\in[0,\infty)$.

Clearly, for all $\displaystyle t\in[1,\infty)$, we have $\displaystyle f(t)\leq|\sin(t)|\leq1$.

Thus, if there exists a point $\displaystyle t_{0}\in[0,1]$ such that $\displaystyle f(t_{0})\geq1$ holds, then we can say that $\displaystyle f(t_{0})$ is the maximum value of $\displaystyle f$ on $\displaystyle [0,\infty)$.

By calculating $\displaystyle f^{\prime}(t)=\big(t\cos(t)-\sin(t)\big)/t^{2}$, which implies $\displaystyle f^{\prime}(0)=0$, therefore $\displaystyle f(0)=1\geq1$ is its maximum value.

I saw**mr fantastic**'s arguments after posting the answer.