What is the f(x)=(-2cos(x)-1)/(2*cos^2(x)) ?

The f(x)=(-2cos(x)-1)/(2*cos^2(x)) is

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f(x)=(-2cos(x)-1)/(2*cos^2(x))

f (x) = \frac{- 2 cos ( x ) - 1}{2 \cdot cos ^{2} ( x )}

Period : 2 π

Domain : 2 πn \leq x < \frac{π}{2} + 2 πn or \frac{π}{2} + 2 πn < x < \frac{3 π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < x < 2 π + 2 πn

Range : f (x) \leq \frac{1}{2}

X Intercepts : (\frac{2 π}{3} + 2 πn, 0), (\frac{4 π}{3} + 2 πn, 0), Y Intercepts : (0, - \frac{3}{2})

Asymptotes : Vertical x = \frac{3 π}{2} + 2 πn, x = \frac{π}{2} + 2 πn

Extreme Points : Maximum (2 πn, - \frac{3}{2}), Maximum (π + 2 πn, \frac{1}{2})

Interval Notation

Domain : [2 πn, \frac{π}{2} + 2 πn) \cup (\frac{π}{2} + 2 πn, \frac{3 π}{2} + 2 πn) \cup (\frac{3 π}{2} + 2 πn, 2 π + 2 πn)

Range : (- \infty, \frac{1}{2}]

Solution steps

Periodicity of

\frac{- 2 cos ( x ) - 1}{2 \cdot cos ^{2} ( x )} : 2 π

Domain of

\frac{- 2 cos ( x ) - 1}{2 \cdot cos ^{2} ( x )} : 2 πn \leq x < \frac{π}{2} + 2 πn or \frac{π}{2} + 2 πn < x < \frac{3 π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < x < 2 π + 2 πn

Range of

\frac{- 2 cos ( x ) - 1}{2 \cdot cos ^{2} ( x )} : f (x) \leq \frac{1}{2}

Axis interception points of

\frac{- 2 cos ( x ) - 1}{2 \cdot cos ^{2} ( x )} :

X Intercepts

: (\frac{2 π}{3} + 2 πn, 0), (\frac{4 π}{3} + 2 πn, 0),

Y Intercepts

: (0, - \frac{3}{2})

Asymptotes of

\frac{- 2 cos ( x ) - 1}{2 \cdot cos ^{2} ( x )} :

Vertical

: x = \frac{3 π}{2} + 2 πn, x = \frac{π}{2} + 2 πn

Extreme Points of

\frac{- 2 cos ( x ) - 1}{2 \cdot cos ^{2} ( x )} :

Maximum

(2 πn, - \frac{3}{2}),

Maximum

(π + 2 πn, \frac{1}{2})

f (x) = - 5 x^{2} + 2 x - 3

f (x) = \frac{- 1}{x ^{2} - 36}

f (x) = \frac{( x ^{4} - 1 )}{( x ^{3} - 1 )}

f, f (x_{x_{y_{2} x} x}) = 3 x_{x_{y_{2} x} x}

f (x) = 5 x^{3} - 2 x^{2} + 3 x - 2

What is the f(x)=(-2cos(x)-1)/(2*cos^2(x)) ?
The f(x)=(-2cos(x)-1)/(2*cos^2(x)) is