Given the function g(x)=sec(x) in the interval [-pi/6, pi/3]

We have to find the absolute extrema of the function on the closed interval.

So first let us find the derivative of the function and equate it to zero.

g'(x)=sec(x)tan(x)=0

sec(x) cannot be equal to 0 because its value ranges between...

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Given the function g(x)=sec(x) in the interval [-pi/6, pi/3]

We have to find the absolute extrema of the function on the closed interval.

So first let us find the derivative of the function and equate it to zero.

g'(x)=sec(x)tan(x)=0

sec(x) cannot be equal to 0 because its value ranges between 1 to +infinity and -1 to infinity only.

Therefore, tan(x)=0 which implies,

x=``

So here the critical points are x=-pi/6, 0, pi/3 which falls in the interval

[-pi/6, pi/3].

So now substituting the critical points in the given function we get the absolute extremas.

g(0)=sec(0)=1

g(-pi/6)=sec(-pi/6)=`2/sqrt(3)`

g(pi/3)=sec(pi/3)=2

So the absolute maximum value is g(x)=2 at x=pi/3(critical as well as end point) and the absolute minimum value is g(x)=1 at x=0 (critical point)