Question

Verify Rolle’s theorem for the function f(x)  `(2)/(e^x + e^-x)` on [– 1, 1].

Answer 1

The function e^x, e^–x and 2  are continuous and differentiable on their respective domains.

∴ f(x) = `(2)/(e^x + e^-x)` is continuous on [– 1, 1] and differentiating on (–1, 1), because e^x + e^–x ≠ 0 x ∈ [– 1, 1].

Now f(– 1) = `(2)/(e^-1 + e) = (2)/(e + e^-1)`
and
f(1) = `(2)/(e + e^-1)`
∴ f(-1) = f(1)
Thus, the function f satisfies all the conditions of the Rolle’s theorem.
∴ there exist c ∈ (–1, 1) such that f'(c) = 0

Now, f(x) = `(2)/(e^x + e^-x)`

∴ f(x) =  `d/dx(2/(e^x + e^-x))`

= `2"d"/dx(e^x + e^-x)^`

= `2( -1)(e^x + e^-x)^-2.d/dx(e^x + e^-x)`

= `(-2)/(e^x + e^-x)^2 xx [e^x + e^-x - (- 1)]`

= `(-2(e^x - e^-x))/(e^x + e^-x)^2`

∴  f'(c) = `(-2(e^c - e^-c))/(e^c + e^-c)^2`

∴ f'(c) = `0 "gives", (-2(e^c - e^-c))/(e^c + e^-c)^2` = 0

∴ e^c – e^{– c} = 0

∴ e^c = ^{– c} = `(1)/e^c`

∴ e^2c = 1 = e°

∴ 2c = 0
∴ c = 0 ∈ (–1, 1)
Hence, Rolle's theorem is verified.