Question

Verify Rolle's theorem for the following function on the indicated interval f (x) = \[{e^{1 - x}}^2\] on [−1, 1] ?

Answer 1

The given function is \[f\left( x \right) = e^{1 - x^2}\] . 

Since exponential function  is everywhere continuous and differentiable, \[e^{1 - x^2}\] 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, \[f\left( x \right)\] satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists \[c \in \left( - 1, 1 \right)\] such that \[f'\left( c \right) = 0\] .

We have

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

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

Thus,

\[c = 0 \in \left( - 1, 1 \right)\] such that 

\[f'\left( c \right) = 0\] .

​Hence, Rolle's theorem is verified.