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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).
Solution
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).
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