`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.

`f(x)=sec(x)`

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)`

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

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

`0=sec(x)tan(x)`

Then, set each factor equal to zero

`secx=0`

`x= {O/ }`    

(There are...

`f(x)=sec(x)`

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)`

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

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

`0=sec(x)tan(x)`

Then, set each factor equal to zero

`secx=0`

`x= {O/ }`    

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

`tanx=0`

`x={0,pi,2pi,...pik}`

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)`

`x=pi/6`

`f'(x)=sec(pi/6)tan(pi/6)=(2sqrt3)/3*sqrt3/3=(2*3)/3=2/3`

`x=pi/4`

`f'(x)=sec(pi/4)tan(pi/4)=sqrt2*1=sqrt2`

`x=pi/3`

`f'(x)=sec(pi/3)tan(pi/3)=2*sqrt3=2sqrt3`

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)` .