So far, no one understands how to solve the boundedness of judgment function. Who knows how to solve the following problem? There must be concrete steps. Thank you very much

Strictly follow the definition of the textbook! Did you study advanced mathematics or mathematical analysis? I don't remember the definition in the textbook very well! I can only tell you the general process, but you have to turn to the book to add all the points!

Solution: (1) y = 3+2cosx;

For any given x, there is M=5, so y = 3+2cosx.

For any given x, there is N= 1, so y = 3+2cosx >: = 1 is a constant, so y = 3+2cosx has a lower bound.

To sum up, y=3+2cosx is bounded!

(2)y=2x^2+ 1；

For any large real number M> 1/2, x0=M exists.

So y = 2× 0 2+1-m = m (2m-1)+1> 0.

That is to say, for any large real number m, x0 exists, making y >: M, so there is no upper bound for y.

For any x, there is M= 1, so y = 2x 2+ 1 >: =M, so y has a lower bound.

Functions with upper and lower bounds are unbounded functions, so y = 2x 2+ 1 is unbounded.

(3)y = 1+arctanx；

unbounded function

(4) y=3sin2x-5cos3x

bounded function

Brother, I'll tell you how. In fact, it is similar to the truth of high school maximum. The function has a maximum value and a minimum value (infinitely close to a certain value) in the definition domain, which is bounded!

University studies pay attention to format specification, in fact, the method is still used in high school!