Question

Find a point on the curve y = x² + x, where the tangent is parallel to the chord joining (0, 0) and (1, 2) ?

Answer 1

​Let:

\[f\left( x \right) = x^2 + x\] 

The tangent to the curve is parallel to the chord joining the points \[\left( 0, 0 \right)\] and \[\left( 1, 2 \right)\] .

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

\[\therefore\] \[a = 0, b = 1\]

The polynomial function is everywhere continuous and differentiable.
So, \[f\left( x \right) = x^2 + x\] is continuous on \[\left[ 0, 1 \right]\] and differentiable on \[\left( 0, 1 \right)\] .

Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently, there exists  \[c \in \left( 0, 1 \right)\] such that 

\[f'\left( c \right) = \frac{f\left( 1 \right) - f\left( 0 \right)}{1 - 0}\].

Now, 

\[f\left( x \right) = x^2 + x\]\[\Rightarrow\] \[f'\left( x \right) = 2x + 1\],\[f\left( 1 \right) = 2, f\left( 0 \right) = 0\]

\[\therefore\] \[f'\left( x \right) = \frac{f\left( 1 \right) - f\left( 0 \right)}{1 - 0}\]\[\Rightarrow\]  \[2x + 1 = \frac{2 - 0}{1 - 0} \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2}\]

Thus, \[c = \frac{1}{2} \in \left( 0, 1 \right)\] such that ​\[f'\left( c \right) = \frac{f\left( 1 \right) - f\left( 0 \right)}{1 - 0}\] .

Clearly,

\[f\left( c \right) = \left( \frac{1}{2} \right)^2 + \frac{1}{2} = \frac{3}{4}\].

Thus, \[\left( c, f\left( c \right) \right)\] i.e.​  

\[\left( \frac{1}{2}, \frac{3}{4} \right)\] is a point on the given curve where the tangent is parallel to the chord joining the points (4, 0) and (5, 1).