Question

Let f be a function defined on [a, b] such that f '(x) > 0, for all x ∈ (a, b). Then prove that f is an increasing function on (a, b).

Answer 1

Let x₁, x₂, ∈ (a, b) such that x₁ < x₂ ∈ f (x) is differentiable on (a, b) and [x₁, x₂] ⊂ (a, b)

∴ f(x) is continuous on [x₁, x₂] and differentiable on (x₁, x₂).

∴ According to Lagrange mean theorem,

Here there exists c ∈ (x₁, x₂) such that

`f'(c) = (f(x_2) - f(x_1))/(x_2 - x_1)`           ...(1)

Since for all x ∈ (a, b), f'(x) > 0

∴ In particular, f'(c) > 0

Now, f'(c) > 0 `=> (f(x_2) - f(x_1))/(x_2 - x_1) > 0`

⇒ f(x₂) - f(x₁) > 0       ...[∵ x₂ - x₁ > 0 when x₁ - x₂]

⇒ f(x₂) > f(x₁)

⇒ f(x₁) < f(x₂), if x₁ < x₂

Because x₁, x₂ are arbitrary points in (a, b).

∴ x₁ < x₂ 

⇒ f(x₁) < f(x₂) for all

x₁, x₂ ∈ (a, b)

∴ f(x) is increasing in (a, b).