What is the extreme f(x)=2sin(3x)-5 ?

The extreme f(x)=2sin(3x)-5 is Maximum(pi/6+(2pi)/3 n,-3),Minimum(pi/2+(2pi)/3 n,-7)

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extreme f(x)=2sin(3x)-5

extreme points

f (x) = 2 sin (3 x) - 5

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

2 sin (3 x) - 5 : - \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

2 sin (3 x) - 5 \Rightarrow y = - 3

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

Plug the extreme point

x = \frac{π}{2} + \frac{2 π}{3} n

into

2 sin (3 x) - 5 \Rightarrow y = - 7

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

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

(5 - x) (x + 1)^{2}

f (x) = 4 x^{7} + 9 x^{5} - 3 x^{4} + 2 x^{2} - 5

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

f (x) = x^{3} + y^{2} - 6 x^{2} + y - 1

- x

What is the extreme f(x)=2sin(3x)-5 ?
The extreme f(x)=2sin(3x)-5 is Maximum(pi/6+(2pi)/3 n,-3),Minimum(pi/2+(2pi)/3 n,-7)