588. Verify Rolle's Theorem for the function :f(x)= x² - 4 x + 3 in the interval 1 ≤ x ≤ 3.

Here f(x) = x² - 4 x + 3
It is a polynomial in x.
(a) Since every polynomial is a continuous function for every value of x
∴ f(x) is continuous in 1 ≤ x ≤ 3.
(b) f'(x) = 2 x, - 4, which exists in 1 < y < 3
∴ f(x) is derivable in 1 < x < 3.
(c) f(i) = (1)² - 4 × 1 + 3 = 1 - 4 + 3 = 0
f(3) = (3)² - 4 × 3 + 3 = 9 - 12 + 3 = 0
∴ f(1) = f(3)
∴ f(x) 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 < 3.
Now f'(c) = 0 gives us 2 c - 4 = 0 or, c = 2
Now c = 2 lies in the open interval (1, 3)
∴ Rolle's Theorem is verified