Prove that the function f given by f(x) = log sin x is strictly increasing on \((0,\frac{\pi}{2})\) and strictly decreasing on \((\frac{\pi}{2},\pi)\).
Solution and Explanation
We have,
f(x)=log sin x
f'(x)=\(\frac{1}{sinx}\) cosx=cot x
In interval \((0,\frac{\pi}{2})\) f'(x)=cot x>0.
∴ f is strictly increasing in \((0,\frac{\pi}{2})\).
In interval \((\frac{\pi}{2},\pi)\), f'(x)=cot x<0.
∴f is strictly decreasing in \((\frac{\pi}{2},\pi)\).
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Notes on Applications of Derivatives
Concepts Used:
Increasing and Decreasing Functions
Increasing Function:
On an interval I, a function f(x) is said to be increasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) ≤ f(y)
Decreasing Function:
On an interval I, a function f(x) is said to be decreasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) ≥ f(y)
Strictly Increasing Function:
On an interval I, a function f(x) is said to be strictly increasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) < f(y)
Strictly Decreasing Function:
On an interval I, a function f(x) is said to be strictly decreasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) > f(y)
Graphical Representation of Increasing and Decreasing Functions
