What is the extreme f(x)=4cos(3x)+5 ?

The extreme f(x)=4cos(3x)+5 is Maximum((2pi)/3 n,9),Minimum(pi/3+(2pi)/3 n,1)

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extreme f(x)=4cos(3x)+5

extreme points

f (x) = 4 cos (3 x) + 5

Maximum (\frac{2 π}{3} n, 9), Minimum (\frac{π}{3} + \frac{2 π}{3} n, 1)

Solution steps

Find the critical points:

x = \frac{2 π}{3} n, x = \frac{π}{3} + \frac{2 π}{3} n

Domain of

4 cos (3 x) + 5 : - \infty < x < \infty

Find intervals:

Decreasing

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

Increasing

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

Plug the extreme point

x = \frac{2 π}{3} n

into

4 cos (3 x) + 5 \Rightarrow y = 9

Maximum (\frac{2 π}{3} n, 9)

Plug the extreme point

x = \frac{π}{3} + \frac{2 π}{3} n

into

4 cos (3 x) + 5 \Rightarrow y = 1

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

Maximum (\frac{2 π}{3} n, 9), Minimum (\frac{π}{3} + \frac{2 π}{3} n, 1)

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

\frac{x ^{2} - x - 6}{x ^{2} - 7 x + 10}

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

1 - x^{3}

f (t) = 5 u (t - 2)

What is the extreme f(x)=4cos(3x)+5 ?
The extreme f(x)=4cos(3x)+5 is Maximum((2pi)/3 n,9),Minimum(pi/3+(2pi)/3 n,1)