What is the extreme f(x)=3-4sin(3x) ?

The extreme f(x)=3-4sin(3x) is Minimum(pi/6+(2pi)/3 n,-1),Maximum(pi/2+(2pi)/3 n,7)

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extreme f(x)=3-4sin(3x)

extreme points

f (x) = 3 - 4 sin (3 x)

Minimum (\frac{π}{6} + \frac{2 π}{3} n, - 1), Maximum (\frac{π}{2} + \frac{2 π}{3} n, 7)

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

3 - 4 sin (3 x) : - \infty < x < \infty

Find intervals:

Decreasing

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

Increasing

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

Decreasing

: \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

3 - 4 sin (3 x) \Rightarrow y = - 1

Minimum (\frac{π}{6} + \frac{2 π}{3} n, - 1)

Plug the extreme point

x = \frac{π}{2} + \frac{2 π}{3} n

into

3 - 4 sin (3 x) \Rightarrow y = 7

Maximum (\frac{π}{2} + \frac{2 π}{3} n, 7)

Minimum (\frac{π}{6} + \frac{2 π}{3} n, - 1), Maximum (\frac{π}{2} + \frac{2 π}{3} n, 7)

f (x, y) = ln ∣ 4 - x - y ∣

f (x) = (x^{2} - 1)^{4}

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

f (x) = x^{5} - 5 x^{4}

f (x) = x - y

What is the extreme f(x)=3-4sin(3x) ?
The extreme f(x)=3-4sin(3x) is Minimum(pi/6+(2pi)/3 n,-1),Maximum(pi/2+(2pi)/3 n,7)