Single Variable Calculus, Chapter 4, 4.1, Section 4.1, Problem 34

Determine the critical numbers of the function $g(t) = |3t - 4|$

We can rewrite the given function as


$\begin{equation} \begin{aligned} g(t) =& \sqrt{(3t - 4)^2} \\ \\ g'(t) =& \frac{d}{dt} (\sqrt{(3t - 4)^2}) \\ \\ g'(t) =& \frac{d}{dt} [(3t - 4)^2]^{\frac{1}{2}} \\ \\ g'(t) =& \frac{1}{2} [(3t - 4)^2]^{\frac{-1}{2}} \frac{d}{dt} (3t - 4)^2 \\ \\ g'(t) =& \left( \frac{1}{\cancel{2}} \right) [(3t - 4)^2]^{\frac{-1}{2}} (\cancel{2}) (3t-4) \frac{d}{dt} (3t - 4) \\ \\ g'(t) =& [(3t - 4)^2]^{\frac{-1}{2}} (3t - 4) (3) \\ \\ g'(t) =& \frac{3 (3t - 4)}{[(3t - 4)^2]^{\frac{1}{2}}} \end{aligned} \end{equation}$


Solving for critical numbers


$\begin{equation} \begin{aligned} & g'(t) = 0 \\ \\ & 0 = \frac{3 (3t - 4)}{[(3t - 4)^2]^{\frac{-1}{2}}} \\ \\ & \sqrt{(3t - 4)^2} \left[ 0 = \frac{3 (3t - 4)}{\cancel{\sqrt{(3t - 4)^2}}} \right] \cancel{\sqrt{(3t - 4)^2}} \\ \\ & 0 = 3 (3t - 4) \\ \\ & \text{ or } \\ \\ & \frac{\cancel{3} (3t - 4)}{\cancel{3}} = \frac{0}{3} \\ \\ & 3t - 4 = 0 \\ \\ & 3t = 4 \\ \\ & \frac{\cancel{3}t}{\cancel{3}} = \frac{4}{3} \\ \\ & t = \frac{4}{3} \end{aligned} \end{equation}$



Therefore, the critical number is $\displaystyle t = \frac{4}{3}$.