What is the f(x)=((4-2sec^2(x)))/(sec^2(x)) ?

The f(x)=((4-2sec^2(x)))/(sec^2(x)) is

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f(x)=((4-2sec^2(x)))/(sec^2(x))

f (x) = \frac{( 4 - 2 sec ^{2} ( x ) )}{sec ^{2} ( x )}

Period : π

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

Range : - 2 < f (x) \leq 2

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

Asymptotes : None

Extreme Points : Maximum (πn, 2)

Solution steps

Periodicity of

\frac{4 - 2 sec ^{2} ( x )}{sec ^{2} ( x )} : π

Domain of

\frac{4 - 2 sec ^{2} ( x )}{sec ^{2} ( x )} : πn \leq x < \frac{π}{2} + πn or \frac{π}{2} + πn < x < π + πn

Range of

\frac{4 - 2 sec ^{2} ( x )}{sec ^{2} ( x )} : - 2 < f (x) \leq 2

Axis interception points of

\frac{4 - 2 sec ^{2} ( x )}{sec ^{2} ( x )} :

X Intercepts

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

Y Intercepts

: (0, 2)

Asymptotes of

\frac{4 - 2 sec ^{2} ( x )}{sec ^{2} ( x )} :

None

Extreme Points of

\frac{4 - 2 sec ^{2} ( x )}{sec ^{2} ( x )} :

Maximum

(πn, 2)

y = 10 sin (2 t + 3)

y = (2 x^{2} + 24 x - 24) e^{4} - x

f (5) = 2 x + 5

f (\frac{x - 1}{2 x + 3}) = (((x^{2})^{0})^{2})^{2} - 1945 x + 3

y = \frac{1}{( 2 x + 10 )}

What is the f(x)=((4-2sec^2(x)))/(sec^2(x)) ?
The f(x)=((4-2sec^2(x)))/(sec^2(x)) is