Question

Find a point on the curve y = x³ + 1 where the tangent is parallel to the chord joining (1, 2) and (3, 28) ?

Answer 1

​Let : 

\[f\left( x \right) = x^3 + 1\] 

The tangent to the curve is parallel to the chord joining the points  \[\left( 1, 2 \right)\] and \[\left( 3, 28 \right)\] .

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

\[\therefore\] \[a = 1, b = 3\]

The polynomial function is everywhere continuous and differentiable.
So, \[f\left( x \right) = x^3 + 1\] is continuous on \[\left[ 1, 3 \right]\] and differentiable on \[\left( 1, 3 \right)\] .

Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently, there exists \[c \in \left( 1, 3 \right)\] such that 

\[f'\left( c \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{3 - 1}\] .

Now, 

\[f\left( x \right) = x^3 + 1\]\[\Rightarrow\] \[f'\left( x \right) = 3 x^2\] ,\[f\left( 1 \right) = 2, f\left( 3 \right) = 28\]

\[\therefore\] \[f'\left( x \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{3 - 1}\]

\[\Rightarrow\] \[3 x^2 = \frac{26}{2} \Rightarrow 3 x^2 = 13 \Rightarrow x = \pm \sqrt{\frac{13}{3}}\]

Thus,\[c = \sqrt{\frac{13}{3}}\]  such that ​

\[f'\left( c \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{3 - 1}\] .

Clearly, 

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

Thus, \[\left( c, f\left( c \right) \right)\]  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 \[\left( 1, 2 \right)\] and \[\left( 3, 28 \right)\] .