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Question
Prove that the function f(x) = tanx – 4x is strictly decreasing on `((-pi)/3, pi/3)`
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Solution
f(x) = tan x – 4x
⇒ f'(x) = sec2x – 4
When `(-pi)/4 < x < pi/3, 1 < secx < 2`
Therefore, 1 < sec2x < 4
⇒ 3 < (sec2x – 4) < 0
Thus for `(-pi)/4 < x < pi/3`, f'(x) < 0
Hence f is strictly decreasing on `((-pi)/3, pi/3)`.
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