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

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

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

critical points

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) = x^{2} + 2 x + 4

f (x) = \frac{x ^{2}}{x - 9}

(- 2.6) - 3

1 + 3.22 lo g_{34} (x)

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