Question #16211

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Answer 1 · 2017-05-14T15:24:23

Please refer to the Explanation Section.

It follows from the Defn. of the fun. $f, that, 0 < a < b .$ $f," that, "0 lt a lt b.$

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.