What is the extreme f(x)=1+sec^2(x) ?

The extreme f(x)=1+sec^2(x) is Minimum(pin,2)

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extreme f(x)=1+sec^2(x)

extreme points

f (x) = 1 + sec^{2} (x)

Minimum (πn, 2)

Solution steps

Find the critical points:

x = πn

Domain of

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

Find intervals:

Increasing

: πn < x < \frac{π}{2} + πn,

Decreasing

: \frac{π}{2} + πn < x < π + πn

Plug the extreme point

x = πn

into

1 + sec^{2} (x) \Rightarrow y = 2

Minimum (πn, 2)

f (x) = x^{4} - 8 x^{2} + 9

f (x) = ln (5 x^{2} - 4 x - 3)

f (x) = x^{- 2} - 13 x^{- 1}

f (x) = 68 x - x^{3}

f (x) = 2 x^{3} - 7

What is the extreme f(x)=1+sec^2(x) ?
The extreme f(x)=1+sec^2(x) is Minimum(pin,2)