Question

At what point  on the following curve, is the tangent parallel to x-axis y = 12 (x + 1) (x − 2) on [−1, 2] ?

Answer 1

Let \[f\left( x \right) = 12\left( x + 1 \right)\left( x - 2 \right)\]    ...(1)

\[\Rightarrow\] \[f\left( x \right) = 12\left( x^2 - x - 2 \right)\]

\[\Rightarrow\] \[f\left( x \right) = 12 x^2 - 12x - 24\]

Since \[f\left( x \right)\] is a polynomial function, \[f\left( x \right)\] is continuous on \[\left[ - 1, 2 \right]\] and differentiable on \[\left( - 1, 2 \right)\].

Also, \[f\left( 2 \right) = f\left( - 1 \right) = 0\]

Thus, all the conditions of Rolle's theorem are satisfied.
Consequently, there exists at least one point c \[\in \left( - 1, 2 \right)\] for which \[f'\left( c \right) = 0\] .

But \[f'\left( c \right) = 0 \Rightarrow 24c - 12 = 0 \Rightarrow c = \frac{1}{2}\]

\[\therefore f\left( c \right) = f\left( \frac{1}{2} \right) = - 12\left( \frac{3}{2} \right)\left( \frac{3}{2} \right) = - 27\] (using (1))
By the geometrical interpretation of Rolle's theorem, \[\left( \frac{1}{2}, - 27 \right)\] is the point on \[y = 12\left( x + 1 \right)\left( x - 2 \right)\] where the tangent is parallel to the x-axis.