From mean value theorem: $f(b)-f(a)=(b-a) f'\left(x_1\right) ; a<x_1<b$ if $f(x)=\frac{1}{x'}$ then $x_1=$
Answer & explanation
Correct answer: option 1
Since $f(x)=\frac{1}{x}$
$f'\left(x_1\right)=\frac{f(b)-f(a)}{b-a}$
$\Rightarrow -\frac{1}{x_1^2}=\frac{1 / b-1 / a}{b-a}$
$\Rightarrow x_1=\sqrt{a b}$