The Mean Value Theorem statement :-
Let #f# be function : # f : [a,b] rarr RR#.
#(M_1) : # f is continuous on #[a,b]#,
#(M_2) : # f is differentiable on #(a,b)#.
Then, #EE# some #c in (a,b)# such that #f'(c)=(f(b)-f(a))/(b-a)#.
In our case, we find that #f#, being a rational function
# : f : [0,1]rarrRR : f(x)=x/(x+6)#,
is continuous on #[0,1]# and differentiable on #(0,1)#.
Thus the conds. #(M_1) and (M_2)# are satisfied by #f#.
Hence, by M.V.T., #(f(1)-f(0))/(1-0)=1/7=f'(c), where, c in (0,1)#.
But, #f(x)=x/(x+6)=(x+6-6)/(x+6)=(x+6)/(x+6)-1/(x+6)=1-1/(x+6)#
#rarr f'(x)=0-(-1/(x+6)^2)=1/(x+6)^2#.
Therefore, #f'(c)=1/7, c in (0,1) rArr 1/(c+6)^2=1/7 rArr c+6=+-sqrt7#, or,
#c=-6+-sqrt7#
But, because of #c in (0,1), c=-6-sqrt7<0# is not admissible.
#:. c=6-sqrt7~=6-2.646=3.354 in (0,1)#
Enjoy Maths.!
