You are not logged in.
Pages: 1
prove that
sin x≥ 2 x/ pi , x ∈ [0, pi/2]
Offline
First you prove this theorem: If f is a differentiable function and a is a point such that f(a) = 0 and f′(a) > 0, there exists δ > 0 such that f(x) > 0 for all x ∈ (a,a+δ).
Proof:
Using the εδ definition of f′(a) (setting ε = f′(a)), there exists δ > 0 such that ∀x ∈ (a−δ,a+δ)
i.e.
i.e.
If we let x ∈ (a,a+δ). then x−a > 0 and so f(x) > 0.
Last edited by JaneFairfax (2007-12-07 03:44:00)
Offline
Now
Then, by the intermediate-value theorem,
By Rolles theorem,
By Rolles theorem again, this time to f′,
But this is a contradiction as f″(x) = −sin(x) is never 0 for x strictly between 0 and π⁄2.
Therefore it must be that
Last edited by JaneFairfax (2007-12-07 03:46:28)
Offline
Pages: 1