Question

At what point  on the following curve, is the tangent parallel to x-axis y = \[e^{1 - x^2}\] on [−1, 1] ?

Answer 1

\[f\left( x \right) = e^{1 - x^2}\]

Since \[f\left( x \right)\] is an exponential function, which is continuous and derivable on its domain,\[f\left( x \right)\] is continuous on \[\left[ - 1, 1 \right]\] and differentiable on \[\left( - 1, 1 \right)\].

Also, 

\[f\left( 1 \right) = f\left( - 1 \right) = 1\]

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

\[\in \left( - 1, 1 \right)\] for which  \[f'\left( c \right) = 0\] .

But 

\[f'\left( c \right) = 0 \Rightarrow - 2c e^{1 - c^2} = 0 \Rightarrow c = 0 \left( \because e^{1 - c^2} \neq 0 \right)\]

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

By the geometrical interpretation of Rolle's theorem, \[\left( 0, e \right)\] is the point on \[y = e^{1 - x^2}\]  where the tangent is parallel to the x-axis .