Calculus of a Single Variable, Chapter 5, 5.6, Section 5.6, Problem 69

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