Find a point on the curve

Solution:

​Let:

$f(x)=x^{3}+1$

The tangent to the curve is parallel to the chord joining the points $(1,2)$ and $(3,28)$.

Assume that the chord joins the points $(a, f(a))$ and $(b, f(b))$.

$\therefore a=1, b=3$

The polynomial function is everywhere continuous and differentiable.

So, $f(x)=x^{3}+1$ is continuous on $[1,3]$ and differentiable on $(1,3)$.

Thus, both the conditions of Lagrange's theorem are satisfied.

Consequently, there exists $c \in(1,3)$ such that $f^{\prime}(c)=\frac{f(3)-f(1)}{3-1} .$

Now, 

$f(x)=x^{3}+1 \Rightarrow f^{\prime}(x)=3 x^{2}, f(1)=2, f(3)=28$

$\therefore f^{\prime}(x)=\frac{f(3)-f(1)}{3-1} \Rightarrow 3 x^{2}=\frac{26}{2} \Rightarrow 3 x^{2}=13 \Rightarrow x=\pm \sqrt{\frac{13}{3}}$

Thus, $c=\sqrt{\frac{13}{3}}$ such that $f^{\prime}(c)=\frac{f(3)-f(1)}{3-1}$.

Clearly, 

$f(c)=\left[\left(\frac{13}{3}\right)^{\frac{3}{2}}+1\right]$

Thus, $(c, f(c))$, i.e. $\left(\sqrt{\frac{13}{3}}, 1+\left(\frac{13}{3}\right)^{\frac{3}{2}}\right)$, is a point on the given curve where the tangent is parallel to the chord joining the points $(1,2)$ and $(3,28)$.