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
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
The domain of the function is , with .
Let
then
which implies .
Therefore its maximum value is .
On the other hand, attains its inferior value as (as you have shown).
For the second one, let
for (I believe it must be given on the positive halfline).
Clearly, as or , we have (superior value).
And on the other hand, we have
which implies
and thus for all (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.
implies which is the inferior value, because there is no point that holds.
As I said before is unbounded; i.e., its superior value is .
And it takes its minimum value at with .
Note. To find the extreme points of a function , just find the zeros of the equation .
Say is such a point that , then if we say that has a minimum point at , and if we say that has a maximum value at .
PS. I just now saw that it seems hard by proving in the way I told you before.
An obvious solution is x = 0. The function has a removable singularity at x = 0. It makes sense to define f(0) = 1.
Then the global maximum of 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.
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 for .
Clearly, for all , we have .
Thus, if there exists a point such that holds, then we can say that is the maximum value of on .
By calculating , which implies , therefore is its maximum value.
I saw mr fantastic's arguments after posting the answer.