f'(x) = -(1/х²)*cos(1/x)
f'(x) ≥ 0
-(1/х²)*cos(1/x) ≥ 0
-(1/х²)<0 при любом х≠0, поэтому cos(1/x) ≤0
π/2+2πn≤(1/x)≤3π/2+2πn; n∈Z
1/(3π/2+2πn)≤х≤1/(π/2+2πn); п∈z
Длина интервала возрастания 1/(π/2+2πn)-1/(3π/2+2πn)=
2/(π+4πn)-2/(3π+4πn)=(2/π)*((3+4n-1-4n)/((1+4n)*(3+4n)=
(4/π)*(1/((1+4n)*(3+4n))=4/π*(π/4)=1
