Calculus: Early Transcendentals, Chapter 4, 4.1, Section 4.1, Problem 34

The modulus of the equation of the function is defined as it follows:
|3t - 4| = 3t - 4 if 3t - 4>=0 => t >= 4/3
or
|3t - 4| = 4 - 3t if t < 4/3
You need to evaluate the critical numbers of the function, hence, you need to solve t the equation g'(t) = 0, in both cases.
For 't in [4/3,oo)' yields 'g'(t) = (3t - 4)' => g'(t) = 3
Notice that g'(t) != 0 for all values of t in [4/3,oo).
For t in (-oo,4/3) yields g'(t) = (4 - 3t)' => g'(t) = -3
Notice that g'(t) != 0 for all values of t in (-oo,4/3).
Hence, there are no critical numbers for the given function such as g'(t) = 0.