It follows from the Defn. of the fun. #f," that, "0 lt a lt b.#
We know that, #f : [a,b] to RR,# is Continuous on #[a,b]# and
Derivable on #(a,b).#
Also, #f(a)=log((a^2+ab)/((a+b)a))=log1=0,# and,
#f(b)=log((b^2+ab)/((a+b)b))=log1=0,# so that, #f(a)=f(b).#
Thus, all the conditions of Rolle's Theorem are satisfied by the
fun. #f#.
#:. EE" some "c in (a,b), s.t., f'(c)=0.#
#f(x)=log((x^2+ab)/((a+b)x))=log(x^2+ab)-log(a+b)-logx.#
# rArr f'(x)=1/(x^2+ab)*2x-0-1/x=(2x^2-x^2-ab)/(x(x^2+ab)), i.e., #
# f'(x)=(x^2-ab)/(x(x^2+ab))#
#:. f'(c)=0 rArr (c^2-ab)/(c(c^2+ab))=0 rArr c^2=ab rArr c=+-sqrtab.#
# because, c in (a,b), (0 lt a lt b)," we have "c=+sqrtab.#
Hence, the Verification of Rolle's Theorem.
