ts) Consider the function $f(x)=x^{1 / 3}-x^{-2 / 3}$.
a. Determine the critical numbers of $f$.
b. Identify any/all relative extrema of $f$. For this part you only need to identify the $x$-coordinate and the type of relative extrema

Step 1 ：First, we need to find the derivative of the function $f(x)=x^{1 / 3}-x^{-2 / 3}$. The derivative of a function at a certain point gives the slope of the tangent line at that point. Critical numbers of a function are the x-values where the derivative is zero or undefined.

Step 2 ：The derivative of $f(x)$ is $f'(x)=\frac{1}{3}x^{-2/3}-\frac{2}{3}x^{-5/3}$.

Step 3 ：To find the critical numbers, we set the derivative equal to zero and solve for $x$. So, we have $\frac{1}{3}x^{-2/3}-\frac{2}{3}x^{-5/3}=0$.

Step 4 ：Multiplying through by $3x^{5/3}$ to clear the fractions gives $x^{3}-2=0$.

Step 5 ：Solving this equation gives $x=\sqrt[3]{2}$.

Step 6 ：We also need to check where the derivative is undefined. The derivative is undefined when the denominator is zero, which is when $x=0$.

Step 7 ：So, the critical numbers of $f$ are $x=0$ and $x=\sqrt[3]{2}$.

Step 8 ：Next, we need to determine the relative extrema. We do this by using the first derivative test. We pick test points in each interval determined by the critical numbers and evaluate the sign of the derivative at these points.

Step 9 ：Choosing $x=-1, x=1, x=2$ as test points, we find that $f'(-1)>0, f'(1)<0, f'(2)>0$.

Step 10 ：Therefore, by the first derivative test, $f$ has a relative maximum at $x=0$ and a relative minimum at $x=\sqrt[3]{2}$.