What is the critical sin(6x),0<= x<= 2pi ?

The critical sin(6x),0<= x<= 2pi is

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critical sin(6x),0<= x<= 2pi

critical points

sin (6 x), 0 \leq x \leq 2 π

x = 0, x = \frac{π}{12}, x = \frac{π}{4}, x = \frac{5 π}{12}, x = \frac{7 π}{12}, x = \frac{3 π}{4}, x = \frac{11 π}{12}, x = \frac{13 π}{12}, x = \frac{5 π}{4}, x = \frac{17 π}{12}, x = \frac{19 π}{12}, x = 2 π

Solution steps

sin (6 x), 0 \leq x \leq 2 π

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{π}{12}, x = \frac{π}{4}, x = \frac{5 π}{12}, x = \frac{7 π}{12}, x = \frac{3 π}{4}, x = \frac{11 π}{12}, x = \frac{13 π}{12}, x = \frac{5 π}{4}, x = \frac{17 π}{12}, x = \frac{19 π}{12}

Identify critical points not in

f (x)

domain

Domain of

sin (6 x), 0 \leq x \leq 2 π : 0 \leq x \leq 2 π

All critical points are in domain

x = \frac{π}{12}, x = \frac{π}{4}, x = \frac{5 π}{12}, x = \frac{7 π}{12}, x = \frac{3 π}{4}, x = \frac{11 π}{12}, x = \frac{13 π}{12}, x = \frac{5 π}{4}, x = \frac{17 π}{12}, x = \frac{19 π}{12}

Evaluate interval endpoint

x = 0 :

critical point

Evaluate interval endpoint

x = 2 π :

critical point

x = 0, x = \frac{π}{12}, x = \frac{π}{4}, x = \frac{5 π}{12}, x = \frac{7 π}{12}, x = \frac{3 π}{4}, x = \frac{11 π}{12}, x = \frac{13 π}{12}, x = \frac{5 π}{4}, x = \frac{17 π}{12}, x = \frac{19 π}{12}, x = 2 π

y = 6 x - 5

1 + x^{\frac{2}{3}} - x

f (x) = 3 (\frac{1}{4})^{x}

- 3 x^{1 - x} - 2

- 2 x^{3} + 2 x + 1