The problem of maximum and minimum value of mathematical function

f'(x)= 1/( 1+x)-x/2

Let f'(x)=0.

Then 1/( 1+x)=x/2.

x？ +x=2

(x+2)(x- 1)=0

x=-2，x= 1

0 & lt= x & lt= 1， 1/2 & lt； = 1/( 1+x)& lt； = 1，0 & lt； = x/2 & lt； = 1/2

So f'(x)>0, and f(x) is increasing function.

Similarly, 1

So x= 1 is the minimum point, and this is also the minimum value of 1.

The maximum value is at the boundary.

f(0)=0，f(2)= ln3- 1 & gt； 0

f( 1)=ln2- 1/4

So the maximum ln3- 1 and the minimum ln2- 1/4.