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

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

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

x = 0

Domain of

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

Find intervals:

Increasing

: 0 < x < \frac{π}{2}

Plug the extreme point

x = 0

into

sin (x) \Rightarrow y = 0

Minimum (0, 0)

f (x) = 3 - ∣ x - 2 ∣

f (x) = \frac{1}{4} x^{4} - \frac{2}{3} x^{3} - \frac{3}{2} x^{2} + 1

f (x, y) = \frac{3 x ^{2} - 1}{x ^{2} + y ^{2} + 1}

f (x) = 3 x + \frac{27}{x}

f (x) = (x^{2} - 4)^{2} - 4

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