Question

Rolle's Theorem? Values of C?

Answer 1

(1).

Rolle's theorem :

Let a real valued function f (x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b).

If f(a) = f(b), then there exist at least one number c in the open interval (a, b) such that f'(c) = 0.

The function is f(x) = x + 6x^-1 and the interval [1/3, x₁].

f(a) = f(1/3) = 1/3 + 6(1/3)^-1

= 1/3 + 6(3)

= 1/3 + 18

= 55/3.

By the rolles therom, f(a) = f(b).

f(1/3) = f(x₁) = 55/3.

To find value of x₁, substitute x = x₁ in the original function.

f(x₁) = x₁ + 6/x₁

Substitute f(x₁) = 55/3 in the above equation.

55/3 = [(x₁)² + 6]/x₁

55x₁ = 3[(x₁)² + 6]

55x₁ = 3(x₁)² + 18

3(x₁)² - 55x₁ + 18 = 0

3(x₁)² - 54x₁ - x₁ + 18 = 0

3x₁(x₁- 18) - 1(x₁ - 18) = 0

(3x₁ - 1)(x₁ - 18) = 0

3x₁ - 1 = 0 and x₁ - 18 = 0

x₁ = 1/3 and x₁ = 18.

Therefore, value of x₁ = 18.

 

Answer 2

(2).

The function is f(x) = x + 6x^-1 and the interval [1/3, 18].

Apply derivative with respect to x.

f'(x) = 1 - 6x^-2

To find value of c, solve the equation f ' (x) = 0.

f'(x) = 1 - 6x^-2 = 0

1 = 6/x²

x²  = 6

x = ± √6.

The value of x = ± √6 = ± 2.45

Here the negative root is negligible because the value - 2.45 does not lie on the interval [1/3, 18].

So, the value of c = √6 = 2.45.