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).

Solution:

Increasing functions are those functions that increase monotonically within a particular domain,

and decreasing functions are those which decrease monotonically within a particular domain.

Let x₁, x₂ ∈ (a, b) such that x₁ > x₂

Consider the sub-interval [x₁, x₂]

Since f (x) is differentiable on (a, b) and [x₁, x₂] ⊂ (a, b).

Therefore,

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

By the Lagrange's mean value theorem,

there exists c ∈ (x₁, x₂) such that

f (c) = (f (x₂) - f (x₁))/(x₁ - x₂) .... (1)

Since, f' (x) > 0 for all x ∈ (a, b),

so in particular,

f' (c) > 0

⇒ [f (x₂) - f (x₁)] / (x₁ - x₂) > 0 [Using (1)]

⇒ f (x₂) - f (x₁) > 0

⇒ f (x₂) > f (x₁)

⇒ f (x₁) < f (x₂)

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

Therefore, x₁ < x₂

⇒ f (x₁) < f (x₂) for all x₁, x₂ ∈ (a, b)

Hence, f (x) is increasing on (a, b)

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise ME Question 16

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)

Summary:

Given that f be a function defined on [a, b] such that f' (x) > 0 , for all x ∈ (a, b). Hence we have proved that f is an increasing function on (a, b)