What is the extreme f(x)= 7/(4-3sin^2(x)) ?

The extreme f(x)= 7/(4-3sin^2(x)) is Minimum(pin, 7/4),Maximum(pi/2+pin,7)

extreme f(x)= 7/(4-3sin^2(x))

extreme points

f (x) = \frac{7}{4 - 3 sin ^{2} ( x )}

Minimum (πn, \frac{7}{4}), Maximum (\frac{π}{2} + πn, 7)

Solution steps

Find the critical points:

x = πn, x = \frac{π}{2} + πn

Domain of

\frac{7}{4 - 3 sin ^{2} ( x )} : - \infty < x < \infty

Find intervals:

Increasing

: πn < x < \frac{π}{2} + πn,

Decreasing

: \frac{π}{2} + πn < x < π + πn

Plug the extreme point

x = πn

into

\frac{7}{4 - 3 sin ^{2} ( x )} \Rightarrow y = \frac{7}{4}

Minimum (πn, \frac{7}{4})

Plug the extreme point

x = \frac{π}{2} + πn

into

\frac{7}{4 - 3 sin ^{2} ( x )} \Rightarrow y = 7

Maximum (\frac{π}{2} + πn, 7)

Minimum (πn, \frac{7}{4}), Maximum (\frac{π}{2} + πn, 7)

x^{2} - 5, 1 < x < 7

(x - 7) (x - 5)^{2} (x - 3) (x - 1)

\frac{400}{1 + 8 e ^{0.05 x}}

1 + (ln (x))

f (x, y) = 3 x^{2} + 3 x - 2 y + 4 y^{2} - 12 x y

What is the extreme f(x)= 7/(4-3sin^2(x)) ?
The extreme f(x)= 7/(4-3sin^2(x)) is Minimum(pin, 7/4),Maximum(pi/2+pin,7)