question use the first derivative test to find the location of all local extrema for the function given…

Explanation:

Step1: Find the derivative of $f(x)$

First, rewrite $f(x)=-8x^{\frac{5}{3}}+x^{\frac{1}{3}}$. Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, we have $f^\prime(x)=-8\times\frac{5}{3}x^{\frac{5}{3}-1}+\frac{1}{3}x^{\frac{1}{3}-1}=-\frac{40}{3}x^{\frac{2}{3}}+\frac{1}{3}x^{-\frac{2}{3}}=\frac{-40x^{\frac{2}{3}}\times x^{\frac{2}{3}} + 1}{3x^{\frac{2}{3}}}=\frac{-40x^{\frac{4}{3}}+1}{3x^{\frac{2}{3}}}$.

Step2: Find the critical points

Set $f^\prime(x) = 0$, then $-40x^{\frac{4}{3}}+1 = 0$. So $40x^{\frac{4}{3}}=1$, and $x^{\frac{4}{3}}=\frac{1}{40}$. Then $x=\pm(\frac{1}{40})^{\frac{3}{4}}$. Also, $f^\prime(x)$ is undefined when $x = 0$ (since the denominator $3x^{\frac{2}{3}}=0$ at $x = 0$), so the critical points are $x = 0,x=-(\frac{1}{40})^{\frac{3}{4}},x = (\frac{1}{40})^{\frac{3}{4}}$.

Step3: Test the intervals

Choose test points in the intervals $(-\infty,-(\frac{1}{40})^{\frac{3}{4}}),(-(\frac{1}{40})^{\frac{3}{4}},0),(0,(\frac{1}{40})^{\frac{3}{4}}),((\frac{1}{40})^{\frac{3}{4}},\infty)$. For the interval $(-\infty,-(\frac{1}{40})^{\frac{3}{4}})$, let $x=-1$. Then $f^\prime(-1)=\frac{-40\times1 + 1}{3\times1}<0$. For the interval $(-(\frac{1}{40})^{\frac{3}{4}},0)$, let $x=-\frac{1}{8}$. Then $f^\prime(-\frac{1}{8})=\frac{-40\times\frac{1}{16}+1}{3\times\frac{1}{4}}=\frac{-\frac{40}{16}+1}{\frac{3}{4}}=\frac{-\frac{24}{16}}{\frac{3}{4}}=-2<0$. For the interval $(0,(\frac{1}{40})^{\frac{3}{4}})$, let $x=\frac{1}{8}$. Then $f^\prime(\frac{1}{8})=\frac{-40\times\frac{1}{16}+1}{3\times\frac{1}{4}}=-2<0$. For the interval $((\frac{1}{40})^{\frac{3}{4}},\infty)$, let $x = 1$. Then $f^\prime(1)=\frac{-40\times1+1}{3\times1}<0$. Since the sign of $f^\prime(x)$ does not change at the critical points, there are no local maxima or minima.

Answer:

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