What is the extreme 12sin(3x) ?

The extreme 12sin(3x) is Maximum(pi/6+(2pi)/3 n,12),Minimum(pi/2+(2pi)/3 n,-12)

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extreme 12sin(3x)

extreme points

12 sin (3 x)

Maximum (\frac{π}{6} + \frac{2 π}{3} n, 12), Minimum (\frac{π}{2} + \frac{2 π}{3} n, - 12)

Solution steps

Find the critical points:

x = \frac{π}{6} + \frac{2 π}{3} n, x = \frac{π}{2} + \frac{2 π}{3} n

Domain of

12 sin (3 x) : - \infty < x < \infty

Find intervals:

Increasing

: \frac{2 π}{3} n \leq x < \frac{π}{6} + \frac{2 π}{3} n,

Decreasing

: \frac{π}{6} + \frac{2 π}{3} n < x < \frac{π}{2} + \frac{2 π}{3} n,

Increasing

: \frac{π}{2} + \frac{2 π}{3} n < x < \frac{2 π}{3} + \frac{2 π}{3} n

Plug the extreme point

x = \frac{π}{6} + \frac{2 π}{3} n

into

12 sin (3 x) \Rightarrow y = 12

Maximum (\frac{π}{6} + \frac{2 π}{3} n, 12)

Plug the extreme point

x = \frac{π}{2} + \frac{2 π}{3} n

into

12 sin (3 x) \Rightarrow y = - 12

Minimum (\frac{π}{2} + \frac{2 π}{3} n, - 12)

Maximum (\frac{π}{6} + \frac{2 π}{3} n, 12), Minimum (\frac{π}{2} + \frac{2 π}{3} n, - 12)

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

(x + 4)^{3} (3 x - 2)^{2}

f (x) = 3 x^{2} - 12

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

f (x) = 14 x^{2} - 2 x^{3} 2 y^{2} + 4 x y

What is the extreme 12sin(3x) ?
The extreme 12sin(3x) is Maximum(pi/6+(2pi)/3 n,12),Minimum(pi/2+(2pi)/3 n,-12)