#### Question

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

#### Solution

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

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

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 . Is there an error in this question or solution?

Solution for question: At What Point on the Following Curve, is the Tangent Parallel to X-axis Y = E 1 − X 2 on [−1, 1] ? concept: Maximum and Minimum Values of a Function in a Closed Interval. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)