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Q28E
Expert-verifiedFind the critical numbers of the function.
28. \(g(t) = |3t - 4|\)
The is critical point for the function occurs at \(\frac{4}{3}\).
The given function is \(g(t) = |3t - 4|\).
Differentiation is a method of finding the derivative of a function. Differentiation is a process, where we find the instantaneous rate of change in function based on one of its variables.
The given \(g(t)\) is an absolute function. Express the given function as follows.
Obtain the first derivative of the given function \(g(t) = |3t - 4|\).
\(\begin{aligned}{c}g(t) &= \frac{d}{{dt}}\left\{ {\begin{aligned}{*{20}{l}}{3t - 4}&{{\rm{ if }}t \ge \frac{4}{3}}\\{4 - 3t}&{{\rm{ if }}t < \frac{4}{3}}\end{aligned}} \right.\\{g^\prime }(t) &= \left\{ {\begin{aligned}{*{20}{l}}3&{{\rm{ if }}t \ge \frac{4}{3}}\\{ - 3}&{{\rm{ if }}t < \frac{4}{3}}\end{aligned}} \right.\end{aligned}\)
Here, \({g^\prime }(t)\) does not exist at \(t = \frac{4}{3}\) as it cannot be zero ever. Therefore, the only valid critical number occurs at \(t = \frac{4}{3}\) as it satisfies the condition of the definition of critical number. Thus, the only critical number of the function \(g(t) = |3t - 4|\) is \(\frac{4}{3}\).
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