Question

At what point  on the following curve, is the tangent parallel to x-axis y = x² on [−2, 2]
?

Answer 1

Let \[f\left( x \right) = x^2\]

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

Also, \[f\left( 2 \right) = f\left( - 2 \right) = 4\]

Thus, all the conditions of Rolle's theorem are satisfied.
Consequently, there exists at least one point c

\[\in \left( - 2, 2 \right)\] for which  \[f'\left( c \right) = 0\] .

But \[f'\left( c \right) = 0 \Rightarrow 2c = 0 \Rightarrow c = 0\]

\[\therefore f\left( c \right) = f\left( 0 \right) = 0\]

By the geometrical interpretation of Rolle's theorem, \[\left( 0, 0 \right)\] is the point on \[y = x^2\] , where the tangent is parallel to the x-axis.