Using the First Derivative Test to Find Local Extrema

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

Solution 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 \((0,0)\).

286

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.