587. Verify Rolle's Theorem for the function : f(x) = x² - 1 in the interval [- 1, 1]

Let f(x) = x² - 1
It is a polynomial in x
(a)    Since every polynomial is a continuous function for every value of x
∴ f is continuous in [-1, 1]
(b)    f'(x) = 2 x, which exists on (-1, 1)
∴ f is derivable in (-1, 1)
(c)    f(-1) = (-1)² - 1 = 1 - 1 = 0
f(1) = (1)² - 1 = 1 - 1 = 0
∴ f(-1) = f(1)
∴ f satisfies all the conditions of Rolle's Theorem
∴ there must exist at least one value c of x such that f'(c) = 0 where - 1 < c 1.
Now f'(c) = 0 gives us 2 c = 0 or c = 0
Now c = 0 lies in the open interval (-1, 1)
∴ Rolle's Theorem is verified