What is the extreme f(x)=6sin(3x) ?

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

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extreme f(x)=6sin(3x)

extreme points

f (x) = 6 sin (3 x)

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

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

6 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

6 sin (3 x) \Rightarrow y = 6

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

Plug the extreme point

x = \frac{π}{2} + \frac{2 π}{3} n

into

6 sin (3 x) \Rightarrow y = - 6

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

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

f (x) = \frac{e ^{x}}{7 x}

f (x) = 12 + 6 x^{2} - x^{3}

f (x) = \frac{3 x ^{2}}{2}

- x^{2} + 3 x - 5

- x^{4} + x^{2}

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