What is the extreme f(x)=cos(2(x-60))-1 ?

The extreme f(x)=cos(2(x-60))-1 is Maximum(pi/3+pin,0),Minimum((5pi)/6+pin,-2)

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extreme f(x)=cos(2(x-60))-1

extreme points

f (x) = cos (2 (x - 6 0^{\circ})) - 1

Maximum (\frac{π}{3} + πn, 0), Minimum (\frac{5 π}{6} + πn, - 2)

Solution steps

Convert

60

degrees to radians:

\frac{π}{3}

cos (2 (x - \frac{π}{3})) - 1

Find the critical points:

x = \frac{π}{3} + πn, x = \frac{5 π}{6} + πn

Domain of

cos (2 (x - \frac{π}{3})) - 1 : - \infty < x < \infty

Find intervals:

Increasing

: πn \leq x < \frac{π}{3} + πn,

Decreasing

: \frac{π}{3} + πn < x < \frac{5 π}{6} + πn,

Increasing

: \frac{5 π}{6} + πn < x < π + πn

Plug the extreme point

x = \frac{π}{3} + πn

into

cos (2 (x - \frac{π}{3})) - 1 \Rightarrow y = 0

Maximum (\frac{π}{3} + πn, 0)

Plug the extreme point

x = \frac{5 π}{6} + πn

into

cos (2 (x - \frac{π}{3})) - 1 \Rightarrow y = - 2

Minimum (\frac{5 π}{6} + πn, - 2)

Maximum (\frac{π}{3} + πn, 0), Minimum (\frac{5 π}{6} + πn, - 2)

5315

f (x) = e^{12 x} + e^{- x}

f (x) = x^{4} - 72 x^{2} + 1296

f (x) = x e^{\frac{y}{x}}

(x - 2)^{3} (x + 2)

What is the extreme f(x)=cos(2(x-60))-1 ?
The extreme f(x)=cos(2(x-60))-1 is Maximum(pi/3+pin,0),Minimum((5pi)/6+pin,-2)