Prove that the function f given by f(x) = log cos x is strictly decreasing on and strictly increasing on
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question is ----> Prove that the function f given by f(x) = log cos x is strictly decreasing (0,π/2) on and strictly increasing on (π/2 , π).
solution :- given f(x) = log cosx
differentiate f(x) with respect to x
f'(x) =![\frac{1}{cosx}(-sinx) \frac{1}{cosx}(-sinx)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bcosx%7D%28-sinx%29)
= -sinx/cosx = - tanx
e.g., f'(x) = -tanx
we know, any function is strictly increasing only when f'(x) > 0 and function is strictly decreasing only when f'(x) < 0 .
![\underline{ for\:strictly\:increasing,} \underline{ for\:strictly\:increasing,}](https://tex.z-dn.net/?f=%5Cunderline%7B+for%5C%3Astrictly%5C%3Aincreasing%2C%7D)
f'(x) = - tanx > 0
tanx < 0
and we know, tanx will be negative when x belongs to (π/2 , π).
![\underline{for\:strictly\:decreasing ,} \underline{for\:strictly\:decreasing ,}](https://tex.z-dn.net/?f=%5Cunderline%7Bfor%5C%3Astrictly%5C%3Adecreasing+%2C%7D)
f'(x) = - tanx < 0
tanx > 0
and we know , tanx will be positive when x belongs to (0, π/2).
hence, it is clear that f(x) = log cosx is strictly decreasing on (0,π/2) and strictly increasing on (π/2, π).
solution :- given f(x) = log cosx
differentiate f(x) with respect to x
f'(x) =
= -sinx/cosx = - tanx
e.g., f'(x) = -tanx
we know, any function is strictly increasing only when f'(x) > 0 and function is strictly decreasing only when f'(x) < 0 .
f'(x) = - tanx > 0
tanx < 0
and we know, tanx will be negative when x belongs to (π/2 , π).
f'(x) = - tanx < 0
tanx > 0
and we know , tanx will be positive when x belongs to (0, π/2).
hence, it is clear that f(x) = log cosx is strictly decreasing on (0,π/2) and strictly increasing on (π/2, π).
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