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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.
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,
So here the critical points are x=-pi/6, 0, pi/3 which falls in the interval
So now substituting the critical points in the given function we get the absolute extremas.
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)
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