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Question
At what point on the following curve, is the tangent parallel to x-axis y = \[e^{1 - x^2}\] on [−1, 1] ?
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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 \[\left( - 1, 1 \right)\].
Consequently, there exists at least one point c
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