What is the critical f(x)=|sin(x)|,0<x<2pi ?

The critical f(x)=|sin(x)|,0<x<2pi is x=pi+2pin

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critical f(x)=|sin(x)|,0<x<2pi

critical points

f (x) = ∣ sin (x) ∣, 0 < x < 2 π

x = π + 2 πn

Solution steps

∣ sin (x) ∣, 0 < x < 2 π

Absolute value functions are piecewise functions

Find the function intervals:

2 πn < x < π + 2 πn, x = π + 2 πn, π + 2 πn < x < 2 π + 2 πn

Join with the function interval:

2 πn < x < π + 2 πn, x = π + 2 πn, π + 2 πn < x < 2 π + 2 πn

Evaluate the function behavior for each interval

Find where

f^{'} (x)

is equal to zero or undefined

x = π + 2 πn

\frac{2 x ^{2} + 1}{x ^{2} - 1}

2 x^{3} + 12 x^{2} + 18 x

f (x) = 9 x^{2} ln (x)

x^{3} + y^{3} + 3 x y + 5

f (x) = (x + 2 y)^{3} + 3 x - 24 y - 2 (x - 3 y)^{3}

What is the critical f(x)=|sin(x)|,0<x<2pi ?
The critical f(x)=|sin(x)|,0<x<2pi is x=pi+2pin