What is the extreme f(x)=2sec(x),-pi/3 <= x<= pi/3 ?

The extreme f(x)=2sec(x),-pi/3 <= x<= pi/3 is Maximum(-pi/3 ,4),Minimum(0,2),Maximum(pi/3 ,4)

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extreme f(x)=2sec(x),-pi/3 <= x<= pi/3

extreme points

f (x) = 2 sec (x), - \frac{π}{3} \leq x \leq \frac{π}{3}

Maximum (- \frac{π}{3}, 4), Minimum (0, 2), Maximum (\frac{π}{3}, 4)

Solution steps

Find the critical points:

x = - \frac{π}{3}, x = 0, x = \frac{π}{3}

Domain of

2 sec (x), - \frac{π}{3} \leq x \leq \frac{π}{3} : - \frac{π}{3} \leq x \leq \frac{π}{3}

Find intervals:

Decreasing

: - \frac{π}{3} < x < 0,

Increasing

: 0 < x < \frac{π}{3}

Plug the extreme point

x = - \frac{π}{3}

into

2 sec (x) \Rightarrow y = 4

Maximum (- \frac{π}{3}, 4)

Plug the extreme point

x = 0

into

2 sec (x) \Rightarrow y = 2

Minimum (0, 2)

Plug the extreme point

x = \frac{π}{3}

into

2 sec (x) \Rightarrow y = 4

Maximum (\frac{π}{3}, 4)

Maximum (- \frac{π}{3}, 4), Minimum (0, 2), Maximum (\frac{π}{3}, 4)

f (x) = - 3 x^{3} + 3 x^{2} - x - 1

- x^{3} - 6 x^{2}

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

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

f (x) = sin^{2} (x), 0 \leq x \leq \frac{2 π}{3}

What is the extreme f(x)=2sec(x),-pi/3 <= x<= pi/3 ?
The extreme f(x)=2sec(x),-pi/3 <= x<= pi/3 is Maximum(-pi/3 ,4),Minimum(0,2),Maximum(pi/3 ,4)