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

The extreme 5sin(3x)-5 is Maximum(pi/6+(2pi)/3 n,0),Minimum(pi/2+(2pi)/3 n,-10)

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extreme 5sin(3x)-5

extreme points

5 sin (3 x) - 5

Maximum (\frac{π}{6} + \frac{2 π}{3} n, 0), Minimum (\frac{π}{2} + \frac{2 π}{3} n, - 10)

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

5 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

5 sin (3 x) - 5 \Rightarrow y = 0

Maximum (\frac{π}{6} + \frac{2 π}{3} n, 0)

Plug the extreme point

x = \frac{π}{2} + \frac{2 π}{3} n

into

5 sin (3 x) - 5 \Rightarrow y = - 10

Minimum (\frac{π}{2} + \frac{2 π}{3} n, - 10)

Maximum (\frac{π}{6} + \frac{2 π}{3} n, 0), Minimum (\frac{π}{2} + \frac{2 π}{3} n, - 10)

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

\frac{1}{2} (5 x - 3)

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

f (x, y) = x^{2} + 4 x y + 5 y^{2} - 7 x + 9 y - 10

f (x) = 9 x - 18 cos (x), - 2 \leq x \leq 0

What is the extreme 5sin(3x)-5 ?
The extreme 5sin(3x)-5 is Maximum(pi/6+(2pi)/3 n,0),Minimum(pi/2+(2pi)/3 n,-10)