Calculus of a Single Variable, Chapter 5, 5.3, Section 5.3, Problem 27

f(x)=ln(x-3)
Take note that a function is strictly monotonic if it is increasing on its entire domain or decreasing on its entire domain.
For our function,
f(x)=ln(x-3)
to determine if it is strictly monotonic, let's first figure out its domain.
Take note that in logarithm, its argument should be above zero. So to get its domain, set its argument x-3 greater than zero.
x-3gt0
xgt3
So the domain of the given function is (3, oo) .
Then, let's apply the derivative. It will be strictly monotonic if there is no sign change in the value of f'(x).
The derivative of the function
f(x) = ln(x-3)
is
f'(x) = 1/(x-3)
Notice that the derivative of the function can never be zero. Because of that, the function f(x) has no critical numbers. This means that there will be no sign change in the value of f'(x). To verify, assign values to x falls within the domain of the function and plug-in them to f'(x).
x=4
f'(x) = 1/(4-3)=1
x=10
f'(x) = 1/(10-3)=1/7
x=23
f'(x) =1/(23-3)=1/20
x=42
f'(x)=1/(42-3)=1/39
Notice that on the interval (3,oo) , the value of f'(x) is always positive. There is no sign change in the value of f'(x). So the function is entirely increasing on its domain.
Therefore, the function f(x)=ln(x-3) is strictly monotonic on its entire domain.