What is the extreme f(t)=9cos(2t) ?

The extreme f(t)=9cos(2t) is Maximum(pin,9),Minimum(pi/2+pin,-9)

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extreme f(t)=9cos(2t)

extreme points

f (t) = 9 cos (2 t)

Maximum (πn, 9), Minimum (\frac{π}{2} + πn, - 9)

Solution steps

Find the critical points:

t = πn, t = \frac{π}{2} + πn

Domain of

9 cos (2 t) : - \infty < t < \infty

Find intervals:

Decreasing

: πn < t < \frac{π}{2} + πn,

Increasing

: \frac{π}{2} + πn < t < π + πn

Plug the extreme point

x = πn

into

9 cos (2 t) \Rightarrow y = 9

Maximum (πn, 9)

Plug the extreme point

x = \frac{π}{2} + πn

into

9 cos (2 t) \Rightarrow y = - 9

Minimum (\frac{π}{2} + πn, - 9)

Maximum (πn, 9), Minimum (\frac{π}{2} + πn, - 9)

S (X, Y) = X Y

f (x) = 2 x^{3} - 36 x^{2} + 162 x - 4

f (x) = 20 x (1 - x)^{3}

4 x^{3} - 48 x - 1

f (x) = x^{3} - 9 x^{2} + 15 x + 8

What is the extreme f(t)=9cos(2t) ?
The extreme f(t)=9cos(2t) is Maximum(pin,9),Minimum(pi/2+pin,-9)