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 $\displaystyle t = \frac{4}{3}$.