What is the extreme y=4cos(x/2)+2 ?

The extreme y=4cos(x/2)+2 is Maximum(4pin,6),Minimum(2pi+4pin,-2)

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extreme y=4cos(x/2)+2

extreme points

y = 4 cos (\frac{x}{2}) + 2

Maximum (4 πn, 6), Minimum (2 π + 4 πn, - 2)

Solution steps

Find the critical points:

x = 4 πn, x = 2 π + 4 πn

Domain of

4 cos (\frac{x}{2}) + 2 : - \infty < x < \infty

Find intervals:

Decreasing

: 4 πn < x < 2 π + 4 πn,

Increasing

: 2 π + 4 πn < x < 4 π + 4 πn

Plug the extreme point

x = 4 πn

into

4 cos (\frac{x}{2}) + 2 \Rightarrow y = 6

Maximum (4 πn, 6)

Plug the extreme point

x = 2 π + 4 πn

into

4 cos (\frac{x}{2}) + 2 \Rightarrow y = - 2

Minimum (2 π + 4 πn, - 2)

Maximum (4 πn, 6), Minimum (2 π + 4 πn, - 2)

K (r, s) = 5 r - s

f (x) = sin (x) cos (x), 0 \leq x \leq 2 π

f (x) = 20 x^{3} - 3 x^{5}

f (x, y) = 9 x^{2} + y^{2}

f (x) = \frac{x ^{3}}{3} - x^{2} - 15 x

What is the extreme y=4cos(x/2)+2 ?
The extreme y=4cos(x/2)+2 is Maximum(4pin,6),Minimum(2pi+4pin,-2)