The constant c of Lagrange’s theorem for $f(x)=\frac{x}{x-1}$ in [2, 4] is:

$\sqrt{3}+1$

f (x ) is continuous on [2, 4] and derivable on (2, 4).

Also, $f'(x)=\frac{(x-1)1-x(1)}{(x-1)^2}=-\frac{1}{(x-1)^2}$

∴ By Lagrange’s mean value theorem

$∀\, c ∈ (2, 4)$ such that $f'(c)=\frac{f(4)-f(2)}{4-2}⇒\frac{-1}{(c-1)^2}=\frac{4/3-2/1}{2}$

$⇒(c-1)^2=3⇒c=1±\sqrt{3}⇒c=\sqrt{3}+1$