$f(x) = x^3 \ in \ the \ interval \ [1 , 3]$ $...........................(1)$
1. $f(x)$ being polynomial is continuous in $[1,3]$
2. $f(x)$ being polynomial is differentiable in $[1,3]$
Both are satisfying to apply Lagrange's mean value theorem. According to this there exists a point $c$ in $[1,3]$ such that
$f'(c) = \underline{f(b)-f(a)}$
$b - a$
$f(3) = 3^3 = 27$
$f(1) = 1^3 = 1$
differentiating w.r.t to $x$, we get
$f'(x) = 3x^2$
$f'(c) = 3c^2$
$3c^2 = (27 - 1) / (3 - 1)$
$3c^2 = 26 / 2$
$c = \pm \sqrt{13*3/3*3}$
$c = \pm \sqrt{13*3}/3$
$c = \pm \sqrt{39}/3$
As $c \ \varepsilon [1 , 3]$
So $c = \sqrt{39}/3$
$x = \sqrt{39}/3$
From $(1) , y = (13\sqrt{39})/3$
$\therefore \bigl(\begin{smallmatrix} \sqrt{39}/3, & (13\sqrt{39})/3 \end{smallmatrix}\bigr)$ is the point on the given curve
$f(x) =x^3$, where tangent is parallel to the chord joining $(1,1)$ and $(3,27)$