What is the extreme y=2x+sin(4x),-pi/3 <= x<= pi/3 ?

The extreme y=2x+sin(4x),-pi/3 <= x<= pi/3 is

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extreme y=2x+sin(4x),-pi/3 <= x<= pi/3

extreme points

y = 2 x + sin (4 x), - \frac{π}{3} \leq x \leq \frac{π}{3}

Maximum (- \frac{π}{3}, - \frac{2 π}{3} + \frac{3}{2}), Minimum (- \frac{π}{6}, - \frac{π}{3} - \frac{3}{2}), Maximum (\frac{π}{6}, \frac{π}{3} + \frac{3}{2}), Minimum (\frac{π}{3}, \frac{2 π}{3} - \frac{3}{2})

Solution steps

Find the critical points:

x = - \frac{π}{3}, x = - \frac{π}{6}, x = \frac{π}{6}, x = \frac{π}{3}

Domain of

2 x + sin (4 x), - \frac{π}{3} \leq x \leq \frac{π}{3} : - \frac{π}{3} \leq x \leq \frac{π}{3}

Find intervals:

Decreasing

: - \frac{π}{3} < x < - \frac{π}{6},

Increasing

: - \frac{π}{6} < x < \frac{π}{6},

Decreasing

: \frac{π}{6} < x < \frac{π}{3}

Plug the extreme point

x = - \frac{π}{3}

into

2 x + sin (4 x) \Rightarrow y = - \frac{2 π}{3} + \frac{3}{2}

Maximum (- \frac{π}{3}, - \frac{2 π}{3} + \frac{3}{2})

Plug the extreme point

x = - \frac{π}{6}

into

2 x + sin (4 x) \Rightarrow y = - \frac{π}{3} - \frac{3}{2}

Minimum (- \frac{π}{6}, - \frac{π}{3} - \frac{3}{2})

Plug the extreme point

x = \frac{π}{6}

into

2 x + sin (4 x) \Rightarrow y = \frac{π}{3} + \frac{3}{2}

Maximum (\frac{π}{6}, \frac{π}{3} + \frac{3}{2})

Plug the extreme point

x = \frac{π}{3}

into

2 x + sin (4 x) \Rightarrow y = \frac{2 π}{3} - \frac{3}{2}

Minimum (\frac{π}{3}, \frac{2 π}{3} - \frac{3}{2})

Maximum (- \frac{π}{3}, - \frac{2 π}{3} + \frac{3}{2}), Minimum (- \frac{π}{6}, - \frac{π}{3} - \frac{3}{2}), Maximum (\frac{π}{6}, \frac{π}{3} + \frac{3}{2}), Minimum (\frac{π}{3}, \frac{2 π}{3} - \frac{3}{2})

x^{2} + 10 x + 24

x^{2} + 4 x - 1

f (x) = 2 x^{3} - 6 x + 3, - 2 \leq x \leq 2

f (x) = - 20 x^{5} - 5 x^{4} + x^{2} - 17

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 8, - 2 \leq x \leq 4

What is the extreme y=2x+sin(4x),-pi/3 <= x<= pi/3 ?
The extreme y=2x+sin(4x),-pi/3 <= x<= pi/3 is