To find the relative extrema of a function, we may apply the First derivative test on a open interval before and after x=c. It states:

a) `f'(x) gt0` from the left and `f'(x)lt0` from right of `x=c` then there is local maximum at `x=c`

b)`f'(x) lt0 ` from the left and `f'(x)gt0` from right of `x=c ` then there is local minimum at `x=c` .

* A sign change of `f'(x)` from left and right of `x=c` will indicate a possible relative extrema (local minima/local maxima). If `f'(x)` has the same sign on both sides of x=c then there is inflection point at `x=c` .

For the given function:`f(x) = arcsec(x)-x` , we have the first derivative:

`f'(x) =1/(sqrt(x^2)sqrt(x^2-1) )-1` .

Equate `f'(x) =0` to solve for `x=c` :

`1/(sqrt(x^2)sqrt(x^2-1) )-1 = 0`

Multiply both sides by` sqrt(x^2)sqrt(x^2-1)` to get `1 -sqrt(x^2)sqrt(x^2-1) = 0`

Move `sqrt(x^2)sqrt(x^2-1) ` to the other side to get `1 = sqrt(x^2)sqrt(x^2-1) `

Square both sides: `1 =x^2 *(x^2-1)`

Expand: `1=x^4-x^2` or `x^4-x^2-1=0`

Apply Quadratic formula:` x= +-sqrt((1+sqrt(5))/2)`

`x= +1.27, -1.27` as the critical values of `f(x)` .

For the critical value x= 1.27, it shows that:

`f'(1.1) = 0.98` which **positive on the left** of `x=1.27`

`f'(1.5) =-0.40` which is **negative on the right** of `x=1.27`

then it follows that **x=1.27** is the location of the **local maximum**

For the critical value `x= -1.27` , it shows that:

`f'(-1.5) =-0.40` which is **negative on the left** of `x=-1.27`

`f'(-1.1) = 0.98` which **positive on the right** of `x=-1.27`

then it follows that** x=-1.27** is the location of the **local minimum**.

With the given function: `f(x)=arcsec(x)-x` , then

**Local maximum**: `f(1.27) = -0.60579`

**Local minimum**: `f(-1.27) = 3.74738`

Please see the attach file for the graph of

`f(x) =arcsec(x)-x`