Matematické Fórum

Skrytý text:
Let us focus on the expression with the absolute value and define



Then it is obvious that .

Let us set



Then the condition is satisfied clearly. Since



and we can restrict just to the interval , thus . Let us define the function as follows



Now we split the solution into 2 cases:

1. Let . Then and



Using derivatives let us find the global maximum of in .



Stationary points can be found from the equations



However, the both equations have no solution, because the expressions on the left side are positive, if . The global maximum can occur also at the endpoints of the interval. Thus and .

2. Let . Then and



In addition,



and the stationary points are given from the pair of the equation mentioned at the end of case 1. Thus



Since



we get that . Hence is a local minimum of a function - it does not solve the problem.

Now let us consider the second equation. It is not necessary to determine , it is enough to find and to solve the problem.



Using simple techniques, we find out that



Therefore there exist four solutions:



Since and then just and satisty the equation . Moreover, the inequalities and imply that and are local maxima of the function .

The global maximum of can occur just at the points , , , . We get that



Thus

Skrytý text:

We use the following identities:



Then we modify the equation:



Now we split the solution into three cases:

1.


2.


3.