Question

Verify the Rolle’s theorem, for the following functions :

(i) f(x) = e^x(sinx - cosx), x ∈ [π/, 5π/4]

(ii) f(x) = (x - a)^m (x - b)ⁿ, x ∈ [a, b], m, n ∈ N

(iii) f(x) = |x|, x ∈ [-1, 1]

Answer 1

(i) Given function

∵ f(x) is polynomial in x.

∴ It is differentiable and continuous every-where.

Hence, f(x) satisfies Rolle’s theorem in interval [π/4, 5π/4]

From(1),

f'(x) = e^x (cos x + sin x) + (sin x – cos x).e^x

f'(x) = e^x (cos x + sin x + sin x – cos x)

Similarly,

e^c 2 sin c = 0

⇒ 2sinc = 0 ⇒ sin c = 0 ⇒ c = π

c = π ∈ (π/4, 5π/4),f(c) = 0

∴ Hence, Rolle’s theorem satisfied.

(ii) f(x) = (x – a)^m (x – b)ⁿ, x ∈ [a, b], m, n ∈ N

Here, (x – a)^m and (x – b)ⁿ both are polynomial. 

On its multiplication a polynomial of power (m + n) is obtained. A polynomial function is continuous everywhere always.

So, f(x) is continuous in [a, b]. Polynomial function is also differentiable.

∴ f'(x) = m(x- a)^m-1 .(x – b)ⁿ + n(x – d)^m (x – b)^n-1

= (x – a)^m-1 (x – b)^n-1 × [m(x – b) + n(x – a)]

= (x – a)^m-1 (x – b)^n-1 × [(m + n) x – mb – na]

Here,f'(x) exists.

∴ f(x) is differentiable in interval (a, b).

Again f(a) = (a – a)^m (a – b)ⁿ = 0

f (b) = (b – a)m (b – b)^m-n = 0

∴ f(a) = f(b) = 0

Hence, Rolle’s theorem satisfies,then in interval (a, b) at least any one point is such that f'(c) = 0.