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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).
Asked by Ravichander Last Modified
Answer
Tasneem
Let x1, x2∈(a,b)x1, x2∈(a,b) such that x1<x2x1<x2.
Consider the sub-interval [x1, x2x1, x2]. Since f (x) is differentiable on (a, b) and [x1, x2]⊂(a,b)[x1, x2]⊂(a,b).
Therefore, f(x) is continous on [x1, x2x1, x2] and differentiable on (x1, x2)(x1, x2).
By the Lagrange's mean value theorm, there exists c∈(x1, x2)c∈(x1, x2) such that
Since f'(x) > 0 for all x∈(a,b)x∈(a,b), so in particular, f'(c) > 0
Since ,
are arbitrary points in (a,b)
Therefore, , for all x ∈ (a, b).
Hence, f (x) is increasing on (a,b).
Tasneem
Since ,
are arbitrary points in (a,b)
Therefore, , for all x ∈ (a, b).
Hence, f (x) is increasing on (a,b).
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