Show that f is strictly monotonic on the given interval and therefore has an inverse function on that interval.

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 , let's take its derivative.

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

Then, set each factor equal to zero

    

(There are...

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 , let's take its derivative.

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

Then, set each factor equal to zero

    

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

So on the interval , 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).

Notice that on the interval , 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 is strictly monotonic on the interval .