What is the extreme f(x)=cos(3x),-1/6 pi<x< 1/2 pi ?

The extreme f(x)=cos(3x),-1/6 pi<x< 1/2 pi is Maximum(0,1),Minimum(pi/3 ,-1)

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extreme f(x)=cos(3x),-1/6 pi<x< 1/2 pi

extreme points

f (x) = cos (3 x), - \frac{1}{6} π < x < \frac{1}{2} π

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

Solution steps

Find the critical points:

x = 0, x = \frac{π}{3}

Domain of

cos (3 x), - \frac{1}{6} π < x < \frac{1}{2} π : - \frac{1}{6} π < x < \frac{1}{2} π

Find intervals:

Increasing

: - \frac{1}{6} π < x < 0,

Decreasing

: 0 < x < \frac{π}{3},

Increasing

: \frac{π}{3} < x < \frac{1}{2} π

Plug the extreme point

x = 0

into

cos (3 x) \Rightarrow y = 1

Maximum (0, 1)

Plug the extreme point

x = \frac{π}{3}

into

cos (3 x) \Rightarrow y = - 1

Minimum (\frac{π}{3}, - 1)

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

f (x) = 8 x^{3}

f (x y) = 2 x y + 80 y - 15 y^{2} - 0.1 x^{2} y - 120

f (x) = x^{3} - 2 x^{2} - 5 x + 6

f (x) = x^{4} + 2 x^{3} - 7

f (x) = 8 x^{\frac{1}{3}} - x^{\frac{4}{3}}, - 8 \leq x \leq 8

What is the extreme f(x)=cos(3x),-1/6 pi<x< 1/2 pi ?
The extreme f(x)=cos(3x),-1/6 pi<x< 1/2 pi is Maximum(0,1),Minimum(pi/3 ,-1)