f(x)=1/x,[1/3;1/2]
f`(c)=[m]\frac{f(b)-f(a)}{b-a}[/m]
b=1/2
a=1/3
f(b)=1/(1/2)=2
f(a)=1/(1/3)=3
b-a=1/6
f(b)-f(a)=2-3=-1
[m]\frac{f(b)-f(a)}{b-a}[/m]=-6
f`(x)=-1/x^2
f`(c)=-1/c^2
-1/с^2=-6
c^2=1/6
c=1/sqrt(6)
c=sqrt(6)/6