# Find the points on the curve

Question:

Find the points on the curve $y=x^{3}-3 x$, where the tangent to the curve is parallel to the chord joining $(1,-2)$ and $(2,2)$.

Solution:

Let:

$f(x)=x^{3}-3 x$

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

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

$\therefore a=1, b=2$

The polynomial function is everywhere continuous and differentiable.

So, $f(x)=x^{3}-3 x$ is continuous on $[1,2]$ and differentiable on $(1,2)$.

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

Consequently, there exists $c \in(1,2)$ such that $f^{\prime}(c)=\frac{f(2)-f(1)}{2-1}$.

Now,

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

$\therefore f^{\prime}(x)=\frac{f(2)-f(1)}{2-1} \Rightarrow 3 x^{2}-3=\frac{2+2}{2-1} \Rightarrow 3 x^{2}=7 \Rightarrow x=\pm \sqrt{\frac{7}{3}}$

Thus, $c=\pm \sqrt{\frac{7}{3}}$ such that $f^{\prime}(c)=\frac{f(2)-f(1)}{2-1}$.

Clearly,

$f\left(\sqrt{\frac{7}{3}}\right)=\left[\left(\frac{7}{3}\right)^{\frac{3}{2}}-3 \sqrt{\frac{7}{3}}\right]=\sqrt{\frac{7}{3}}\left[\frac{7}{3}-3\right]=\sqrt{\frac{7}{3}}\left[\frac{-2}{3}\right]=\frac{-2}{3} \sqrt{\frac{7}{3}}$ and $f\left(-\sqrt{\frac{7}{3}}\right)=\frac{2}{3} \sqrt{\frac{7}{3}}$

$\therefore f(c)=\mp \frac{2}{3} \sqrt{\frac{7}{3}}$

Thus, $(c, f(c))$, i.e. $\left(\pm \sqrt{\frac{7}{3}}, \mp \frac{2}{3} \sqrt{\frac{7}{3}}\right)$, are points on the given curve where the tangent is parallel to the chord joining the points $(1,-2)$ and $(2,2)$.