f'=4x^3- 12x^2=4x^2(x-3)

F'=0, so x=0 and x=3.

X≥3, f'≥0, increasing function; X≤3, f'≤0, negative function.

4:x≤0，f '(x)= 2x-2 = 2(x- 1)& lt； 0, negative function

x & gt0,f'(x)=6x^2- 18x+ 12=6(x^2-3x+2)=6(x- 1)(x-2)

X=0~ 1, f'≥0, increasing function;

X= 1~2, f'≤0, negative function.

X>2, f'≥0, increasing function;

5:( 1)sin(x/2)cos(y/2)+cos(x/2)sin(y/2)-sin(x/2)cos(x/2)-sin(y/2)cos(y/2)

= cos(x/2)[sin(y/2)-sin(x/2)]+cos(y/2)[sin(x/2)-sin(y/2)]

=[cos(x/2)-cos(y/2)][sin(y/2)-sin(x/2)]

0<x, y< pi?

0 & ltx/2，y/2 & lt； π/2

Sin is increasing function and cos is a decreasing function. Whether x≤y or x≥y, [cos(x/2)-cos(y/2)] and [sin(y/2)-sin(x/2)] are always of the same sign, so the original formula holds.

(2)(tanx)'=(secx)^2>； 0, tanx is an increasing function.

Multiply both sides by 2:

2 arctan[(x+y)/2]≥arctanx+arctany

Both sides have sun drying function.

Then (x+y)/{1-[(x+y)/2] 2} ≥ (x+y)/(1-xy)

x，y & gt0; x+y & gt； 0; About going to x+y

1/{ 1-[(x+y)/2]^2}≥ 1/( 1-xy)

Countdown on both sides:

1-[(x+y)/2]^2≤ 1-xy

xy≤[(x+y)/2]^2=(x^2+2xy+y^2)/4

4xy≤x^2+2xy+y^2

0≤x^2-2xy+y^2=(x-y)^2

Established.

Every step above is reversible, so the original formula holds.