Question

1. a. find the critical number(s) of ( f(x)=9 x^{2 / 3}-3 x ).

b. use the first derivative test to find any local min/max of ( f(x) ).

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 \((8,12)\)