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Q28E

Expert-verifiedFound in: Page 209

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

**Find 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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