`f(x) = sec(x), [0, pi/2)` Show that f is strictly monotonic on the given interval and therefore has an inverse function on that interval.

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Take note that a function is strictly monotonic on a given interval if it is entirely increasing on that interval or entirely decreasing on that interval.

To determine if f(x) is strictly monotonic on the interval `[0, pi/2)` , let's take its derivative.


`f'(x) =sec(x)tan(x)`

Then, determine the critical numbers. To do so, set f'(x) equal to zero.


Then, set each factor equal to zero


`x= {O/ }`    

(There are no angles in which the value of secant will be zero.)



So on the interval `[0,pi/2)` , the only critical number that belongs to it is x=0. Since the critical number is the boundary of the given interval, it indicates that the there is no sign change in the value of f'(x) on [0, pi/2).  To verify, let's assign values to x which falls on that interval and plug-in them to f'(x).

`f'(x) = sec(x)tan(x)`








Notice that on the interval `[0, pi/2)` , the values of f'(x) are all positive. There is no sign change. So the function is entirely increasing on this interval.

Therefore, the function `f(x)=sec(x)` is strictly monotonic on the interval `[0,pi/2)` .

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