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

The extreme f(x)=2sin^2(x),0<= x<= pi is Minimum(0,0),Maximum(pi/2 ,2),Minimum(pi,0)

Popular >

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

extreme points

f (x) = 2 sin^{2} (x), 0 \leq x \leq π

Minimum (0, 0), Maximum (\frac{π}{2}, 2), Minimum (π, 0)

Solution steps

Find the critical points:

x = 0, x = \frac{π}{2}, x = π

Domain of

2 sin^{2} (x), 0 \leq x \leq π : 0 \leq x \leq π

Find intervals:

Increasing

: 0 < x < \frac{π}{2},

Decreasing

: \frac{π}{2} < x < π

Plug the extreme point

x = 0

into

2 sin^{2} (x) \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = \frac{π}{2}

into

2 sin^{2} (x) \Rightarrow y = 2

Maximum (\frac{π}{2}, 2)

Plug the extreme point

x = π

into

2 sin^{2} (x) \Rightarrow y = 0

Minimum (π, 0)

Minimum (0, 0), Maximum (\frac{π}{2}, 2), Minimum (π, 0)

f (x) = 4 cos (3 x) + 5, (0, 2 π)

f (x, y) = 14 x y - x^{3} - 7 y^{2}

f (x, y) = x^{2} + x y + y^{2} - 25 y + 208

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

y = \frac{x ( z + 2 )}{7}

What is the extreme f(x)=2sin^2(x),0<= x<= pi ?
The extreme f(x)=2sin^2(x),0<= x<= pi is Minimum(0,0),Maximum(pi/2 ,2),Minimum(pi,0)