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 

\[f'\left( c \right) = \frac{f\left( 2 \right) - f\left( 1 \right)}{2 - 1}\] .

Now,

\[f\left( x \right) = x^3 - 3x\]\[\Rightarrow\] \[f'\left( x \right) = 3 x^2 - 3\],\[f\left( 1 \right) = - 2, f\left( 2 \right) = 2\]

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

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

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 \[\left( 2, 2 \right)\] .