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

The extreme sin(2x),0<= x<= pi is Minimum(0,0),Maximum(pi/4 ,1),Minimum((3pi)/4 ,-1),Maximum(pi,0)

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extreme sin(2x),0<= x<= pi

extreme points

sin (2 x), 0 \leq x \leq π

Minimum (0, 0), Maximum (\frac{π}{4}, 1), Minimum (\frac{3 π}{4}, - 1), Maximum (π, 0)

Solution steps

Find the critical points:

x = 0, x = \frac{π}{4}, x = \frac{3 π}{4}, x = π

Domain of

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

Find intervals:

Increasing

: 0 < x < \frac{π}{4},

Decreasing

: \frac{π}{4} < x < \frac{3 π}{4},

Increasing

: \frac{3 π}{4} < x < π

Plug the extreme point

x = 0

into

sin (2 x) \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = \frac{π}{4}

into

sin (2 x) \Rightarrow y = 1

Maximum (\frac{π}{4}, 1)

Plug the extreme point

x = \frac{3 π}{4}

into

sin (2 x) \Rightarrow y = - 1

Minimum (\frac{3 π}{4}, - 1)

Plug the extreme point

x = π

into

sin (2 x) \Rightarrow y = 0

Maximum (π, 0)

Minimum (0, 0), Maximum (\frac{π}{4}, 1), Minimum (\frac{3 π}{4}, - 1), Maximum (π, 0)

f (x) = 5 x + 3 x^{- 1}

f (x, y) = (x^{2} + y^{2}) e^{x^{2} - y^{2}}

f (x) = x^{3} (x - 2)^{2}

x^{3} e^{- x}

f (x, y) = x ln (x^{2} + y^{2})

What is the extreme sin(2x),0<= x<= pi ?
The extreme sin(2x),0<= x<= pi is Minimum(0,0),Maximum(pi/4 ,1),Minimum((3pi)/4 ,-1),Maximum(pi,0)