What is the critical f(x)=16cos(x)+8sin^2(x) ?

The critical f(x)=16cos(x)+8sin^2(x) is x=2pin,x=pi+2pin

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critical f(x)=16cos(x)+8sin^2(x)

critical points

f (x) = 16 cos (x) + 8 sin^{2} (x)

x = 2 πn, x = π + 2 πn

Solution steps

16 cos (x) + 8 sin^{2} (x)

Find where

f^{'} (x)

is equal to zero or undefined

x = 2 πn, x = π + 2 πn

Identify critical points not in

f (x)

domain

Domain of

16 cos (x) + 8 sin^{2} (x) : - \infty < x < \infty

All critical points are in domain

x = 2 πn, x = π + 2 πn

f (x) = \frac{16 x}{( x ^{2} + 4 ) ^{2}}

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

80 x^{3} - 3 x^{5}

2 x - x^{2} - x y

2 x^{3} + x^{2} + 2 x

What is the critical f(x)=16cos(x)+8sin^2(x) ?
The critical f(x)=16cos(x)+8sin^2(x) is x=2pin,x=pi+2pin