What is the extreme f(x)=sin(x)+sin^3(x),-pi<x<pi ?

The extreme f(x)=sin(x)+sin^3(x),-pi<x<pi is Minimum(-pi/2 ,-2),Maximum(pi/2 ,2)

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extreme f(x)=sin(x)+sin^3(x),-pi<x<pi

extreme points

f (x) = sin (x) + sin^{3} (x), - π < x < π

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

Solution steps

Find the critical points:

x = - \frac{π}{2}, x = \frac{π}{2}

Domain of

sin (x) + sin^{3} (x), - π < x < π : - π < x < π

Find intervals:

Decreasing

: - π < x < - \frac{π}{2},

Increasing

: - \frac{π}{2} < x < \frac{π}{2},

Decreasing

: \frac{π}{2} < x < π

Plug the extreme point

x = - \frac{π}{2}

into

sin (x) + sin^{3} (x) \Rightarrow y = - 2

Minimum (- \frac{π}{2}, - 2)

Plug the extreme point

x = \frac{π}{2}

into

sin (x) + sin^{3} (x) \Rightarrow y = 2

Maximum (\frac{π}{2}, 2)

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

y = (x - 1)^{3} (x - 5)

f (x) = 4 x - 6

f (x) = - \frac{2}{x ^{2}}, 0.5 \leq x \leq 2

f (x) = - \frac{2}{x ^{2}}, 0.5 \leq x \leq 5

f (x) = 2 - 3 x

What is the extreme f(x)=sin(x)+sin^3(x),-pi<x<pi ?
The extreme f(x)=sin(x)+sin^3(x),-pi<x<pi is Minimum(-pi/2 ,-2),Maximum(pi/2 ,2)