# Thread: how to find the max,min,sup,inf of these cases..

1. ## how 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

2. The domain of the function $\mathrm{ln}(x)$ is $(0,\infty)$, with $\lim_{x\to0}\mathrm{ln}(x)=-\infty$.
Let
$f(x):=\frac{1}{1+(\mathrm{ln}(x))^{2}},$
then
$f^{\prime}(x):=-\frac{2\mathrm{ln}(x)}{x(1+(\mathrm{ln}(x))^{2})^{ 2}},$
which implies $f^{\prime}(1)=0$.
Therefore its maximum value is $f(1)=1$.
On the other hand, $f$ attains its inferior value as $x\to\infty$ (as you have shown).

For the second one, let
$g(x):=x+\frac{1}{x}$
for $x>0$ (I believe it must be given on the positive halfline).
Clearly, as $x\to0$ or $x\to\infty$, we have $g(x)\to\infty$ (superior value).
And on the other hand, we have
$g^{\prime}(x)=1-\frac{1}{x^{2}},$
which implies $g^{\prime}(1)=0$
and thus $g(x)\geq g(1)=2$ for all $x>0$ (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.

3. so the max and the sup of the first one is 1
what is the min and inf of the first one?

4. $

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

the maximal number is infinty
when x->+infinty

the smallest value is found when
x->-infinity

??

5. 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

6. Originally Posted by transgalactic
so the max and the sup of the first one is 1
what is the min and inf of the first one?
$\lim_{x\to\infty}\ln(x)=\infty$ implies $\lim_{t\to\infty}\frac{1}{1+(\ln(x))^{2}}=0$ which is the inferior value, because there is no $x_{0}$ point that $\frac{1}{1+(\ln(x_{0}))^{2}}=0$ holds.

Originally Posted by transgalactic
$

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

the maximal number is infinty
when x->+infinty

the smallest value is found when
x->-infinity

??
As I said before $x+\frac{1}{x}$ is unbounded; i.e., its superior value is $\infty$.
And it takes its minimum value at $x=1$ with $1+\frac{1}{1}=2$.

Note. To find the extreme points of a function $f$, just find the zeros of the equation $f^{\prime}=0$.
Say $x_{0}$ is such a point that $f^{\prime}(x_{0})=0$, then if $f^{\prime\prime}>0$ we say that $f$ has a minimum point at $x_{0}$, and if $f^{\prime\prime}<0$ we say that $f$ has a maximum value at $x_{0}$.

PS. I just now saw that it seems hard by proving in the way I told you before.

7. in the third case the sinus function
i get
cosx * x -sinx =0

how to solve this equation??

8. Originally Posted by transgalactic
in the third case the sinus function
i get
cosx * x -sinx =0

how to solve this equation??
An obvious solution is x = 0. The function $f(x) = \frac{\sin x}{x}$ has a removable singularity at x = 0. It makes sense to define f(0) = 1.

Then the global maximum of $f(x) = \frac{\sin x}{x}$ 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.

9. Originally Posted by transgalactic
in the third case the sinus function
i get
cosx * x -sinx =0

how to solve this equation??
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 $f(t):=\sin(t)/t$ for $t\in[0,\infty)$.
Clearly, for all $t\in[1,\infty)$, we have $f(t)\leq|\sin(t)|\leq1$.
Thus, if there exists a point $t_{0}\in[0,1]$ such that $f(t_{0})\geq1$ holds, then we can say that $f(t_{0})$ is the maximum value of $f$ on $[0,\infty)$.
By calculating $f^{\prime}(t)=\big(t\cos(t)-\sin(t)\big)/t^{2}$, which implies $f^{\prime}(0)=0$, therefore $f(0)=1\geq1$ is its maximum value.

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