Single Variable Calculus, Chapter 4, 4.1, Section 4.1, Problem 40

Determine the critical numbers of the function $g(x) = x^{\frac{1}{3}} - x^{- \frac{2}{3}}$


$
\begin{equation}
\begin{aligned}

g'(x) =& \frac{d}{dx} (x^{\frac{1}{3}}) - \frac{d}{dx} (x^{- \frac{2}{3}})
\\
\\
g'(x) =& \frac{1}{3} (x)^{- \frac{2}{3}} - \left( \frac{-2}{3} \right) (x)^{- \frac{5}{3}}
\\
\\
g'(x) =& \frac{1}{3(x)^{\frac{2}{3}}} + \frac{2}{3(x)^{\frac{5}{3}}}
\\
\\
g'(x) =& \left[ \frac{1}{3(x)^{\frac{2}{3}}} \cdot \frac{x}{x} \right] + \frac{2}{3(x)^{\frac{5}{3}}}
\\
\\
g'(x) =& \frac{x}{3(x)^{\frac{5}{3}}} + \frac{2}{3 (x)^{\frac{5}{3}}}
\\
\\
g'(x) =& \frac{x + 2}{3 (x)^{\frac{5}{3}}}

\end{aligned}
\end{equation}
$


Solving for critical number


$
\begin{equation}
\begin{aligned}

& g'(x) = 0
\\
\\
& 0 = \frac{x + 2}{3(x)^{\frac{5}{3}}}
\\
\\
& 3(x)^{\frac{5}{3}} \left[ 0 = \frac{x + 2}{\cancel{3 (x)^{\frac{5}{3}}}} \right] \cancel{3 (x)^{\frac{5}{3}}}
\\
\\
& 0 = x+ 2
\\
\\
& 0 - 2 = x + 2 - 2
\\
\\
& -2 = x
\\
\\
& \text{ or }
\\
\\
& x = -2

\end{aligned}
\end{equation}
$


Therefore, the critical number is $x = -2$.