Figure

EXAMPLE 2Using the First Derivative Test to Find Local Extrema

Find the local extrema of $f(x)=x^{2/3}({x-5})$ .

The domain of $f$ is all real numbers and $f$ is continuous on its domain. $\begin{eqnarray*} &&\hspace{1pc} {f^\prime }(x) = x^{2/3}+\left( {\frac{2}{3}}\right) x^{-1/3}({x-5})=\frac{3x+2({x-5})}{3x^{1/3}}=\frac{5}{3}\!\left( {\frac{x-2}{x^{1/3}}}\right)\\[-10.5pt] && \color{#0066A7}{\underset{\hbox{Use the Product Rule.}}{\uparrow}} \end{eqnarray*}$

Since $f^\prime (2) =0$ and $f^\prime (0)$ does not exist, the critical numbers are $0$ and $2$ . The graph of $f$ will have a horizontal tangent line at the point $(2,-3\sqrt[\kern-1pt3\kern1pt]{4})$ and a vertical tangent line at the point $(0,0)$ .

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Table 4 shows the intervals on which $f$ is increasing and decreasing.

Figure 34

$f(x)=x^{2/3}(x-5)$

TABLE 4

Interval	Sign of ${x-2}$	Sign of ${x^{1/3}}$	Sign of ${f^\prime (x) =\dfrac{5}{3}\left( {\dfrac{x-2}{x^{1/3}}}\right) }$	Conclusion
$(-\infty ,0)$	Negative $(-)$	Negative $(-)$	Positive $(+)$	$f$ is increasing
$(0,2)$	Negative $(-)$	Positive $(+)$	Negative $(-)$	$f$ is decreasing
$(2,\infty)$	Positive $(+)$	Positive $(+)$	Positive $(+)$	$f$ is increasing

By the First Derivative Test, $f$ has a local maximum at $0$ and a local minimum at $2$ ; $f( 0) =0$ is a local maximum value and $f(2) =-3\sqrt[3]{4}$ is a local minimum value.