What is the extreme f(x)=24sin(3x) ?

The extreme f(x)=24sin(3x) is Maximum(pi/6+(2pi)/3 n,24),Minimum(pi/2+(2pi)/3 n,-24)

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extreme f(x)=24sin(3x)

extreme points

f (x) = 24 sin (3 x)

Maximum (\frac{π}{6} + \frac{2 π}{3} n, 24), Minimum (\frac{π}{2} + \frac{2 π}{3} n, - 24)

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

24 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

24 sin (3 x) \Rightarrow y = 24

Maximum (\frac{π}{6} + \frac{2 π}{3} n, 24)

Plug the extreme point

x = \frac{π}{2} + \frac{2 π}{3} n

into

24 sin (3 x) \Rightarrow y = - 24

Minimum (\frac{π}{2} + \frac{2 π}{3} n, - 24)

Maximum (\frac{π}{6} + \frac{2 π}{3} n, 24), Minimum (\frac{π}{2} + \frac{2 π}{3} n, - 24)

y = 0.25 x^{4} - (\frac{1}{3}) x^{3}

6 x

f (x, y) = 3 x^{2} + 2 y^{2} - 6 x + 8 y

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

e^{3 x^{2}}

What is the extreme f(x)=24sin(3x) ?
The extreme f(x)=24sin(3x) is Maximum(pi/6+(2pi)/3 n,24),Minimum(pi/2+(2pi)/3 n,-24)