Question

Find the points on the curve y = x³ − 3x, where the tangent to the curve is parallel to the chord joining (1, −2) and (2, 2).

Answer 1

​Let:
$f\left(x\right)=x^3-3x$

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

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

$\therefore$ $a=1,b=2$

The polynomial function is everywhere continuous and differentiable.

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

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

Consequently, there exists $c\in \left(1,2\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$.

Now,
$f\left(x\right)=x^3-3x$$\Rightarrow$$f\text{'}\left(x\right)=3x^2-3$, $f\left(1\right)=-2,f\left(2\right)=2$

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

Thus, $c=\pm \sqrt{\frac{7}{3}}$ such that ​$f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$.

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

∴ $f\left(c\right)=\mp \frac{2}{3}\sqrt{\frac{7}{3}}$

Thus, $\left(c,f\left(c\right)\right)$, 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 $\left(1,-2\right)$ and $\left(2,2\right)$.