3. a. find the critical number(s) of ( f(x)=9 x^{2 / 3}-3 x ).\nb. use the first derivative test to find any…

Explanation:

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

Using the power rule ((x^n)^\prime = nx^{n - 1}), for (f(x)=9x^{\frac{2}{3}}-3x), we have (f^\prime(x)=9\times\frac{2}{3}x^{\frac{2}{3}-1}-3). Simplify (f^\prime(x)=6x^{-\frac{1}{3}}-3=\frac{6}{x^{\frac{1}{3}}}-3=\frac{6 - 3x^{\frac{1}{3}}}{x^{\frac{1}{3}}})

Step2: Find the critical numbers

Set (f^\prime(x) = 0), then (\frac{6 - 3x^{\frac{1}{3}}}{x^{\frac{1}{3}}}=0). The numerator (6 - 3x^{\frac{1}{3}}=0) gives (3x^{\frac{1}{3}}=6), so (x^{\frac{1}{3}} = 2), and (x = 8). Also, (f^\prime(x)) is undefined when (x = 0) (since the denominator (x^{\frac{1}{3}}=0) at (x = 0)).

Step3: First - Derivative Test for (x = 0)

Choose test intervals: ((-\infty,0)) and ((0,8)) and ((8,\infty)) For (x=-1) (in ((-\infty,0))), (f^\prime(-1)=\frac{6}{(-1)^{\frac{1}{3}}}-3=-6 - 3=-9<0) For (x = 1) (in ((0,8))), (f^\prime(1)=\frac{6}{1^{\frac{1}{3}}}-3=6 - 3 = 3>0) Since (f^\prime(x)) changes sign from negative to positive at (x = 0), by the First - Derivative Test, (f(x)) has a local minimum at (x = 0). (f(0)=9\times0^{\frac{2}{3}}-3\times0 = 0)

Step4: First - Derivative Test for (x = 8)

For (x = 1) (in ((0,8))), (f^\prime(1)=3>0) For (x = 27) (in ((8,\infty))), (f^\prime(27)=\frac{6}{27^{\frac{1}{3}}}-3=\frac{6}{3}-3=2 - 3=-1<0) Since (f^\prime(x)) changes sign from positive to negative at (x = 8), by the First - Derivative Test, (f(x)) has a local maximum at (x = 8). (f(8)=9\times8^{\frac{2}{3}}-3\times8=9\times4-24 = 12)

Answer:

A. The critical numbers are (x = 0) and (x = 8). B. Local minimum at ((0,0)) and local maximum at ((8,12))