Verify Roll�s Theorem for the function :f{x) = e^{1 - x²} in [-1, 1]

Solution

Let f(x) = e^{1 - x²}(a)    Since e^{1 - x²} is continuous in [-1, 1]
∴ f is continuous in [-1, 1]
(b)    f'(x) = e^{1 - x²} (- 2 x) = - 2 x e^{1 - x²} , which exists in (-1, 1)
∴ is derivable in (-1, 1)
(c)    f(-1) = e_{1 - 1} = e⁰ = 1
f(1) = e^{1 - 1} = e⁰ = 1
∴ f(-1) = f(1)
∴ satifies all the conditions of Rolle's Theorem.
∴ there must exist at least one value c of x such f' (c) = 0 where - 1 < c < 1
Now f' (c) = 0 gives us - 2 c e^1-c² = 0
⇒ c = 0 ∈ (-1, 1)
∴ Rolle's Theorem is verified.

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Verify Roll�s Theorem for the function :f{x) = e^{1 - x²} in [-1, 1]

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Verify Roll�s Theorem for the function :f{x) = e1 - x2 in [-1, 1]

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Verify Roll�s Theorem for the function :f{x) = e^{1 - x²} in [-1, 1]