To find the critical numbers of the function $f(x) = 2x^{5/3} - 45x^{2/3}$, we need to follow these steps:

Step 1: Take the derivative of the function

We will use the power rule for derivatives to find $f'(x)$.

$f'(x) = \frac{d}{dx} \left( 2x^{5/3} - 45x^{2/3} \right)$

• The derivative of $2x^{5/3}$ is: $\frac{d}{dx} \left( 2x^{5/3} \right) = 2 \cdot \frac{5}{3} x^{(5/3) - 1} = \frac{10}{3} x^{2/3}$

• The derivative of $-45x^{2/3}$ is: $\frac{d}{dx} \left( -45x^{2/3} \right) = -45 \cdot \frac{2}{3} x^{(2/3) - 1} = -30x^{-1/3}$

Thus, the derivative is: $f'(x) = \frac{10}{3} x^{2/3} - 30x^{-1/3}$

Step 2: Set the derivative equal to zero

To find the critical points, set $f'(x) = 0$:

$\frac{10}{3} x^{2/3} - 30x^{-1/3} = 0$

Multiply through by 3 to eliminate the fractions: $10x^{2/3} - 90x^{-1/3} = 0$

Step 3: Factor the equation

Factor out the common term $x^{-1/3}$:

$x^{-1/3} \left( 10x - 90 \right) = 0$

This gives two possibilities:

1. $x^{-1/3} = 0$, but this is not possible since $x^{-1/3}$ is undefined for $x = 0$.
2. $10x - 90 = 0$, solve for $x$: $10x = 90 \quad \Rightarrow \quad x = 9$

Step 4: Analyze undefined points

Since $f'(x)$ contains the term $x^{-1/3}$, the derivative is undefined at $x = 0$. This is also a critical number.

Final Answer:

The critical numbers are $x = 0$ and $x = 9$.

Do you have any questions or need more details? Here are some related questions you might find helpful:

1. How do we identify whether a critical point is a minimum or maximum?
2. What does it mean for a derivative to be undefined at a point?
3. Can a function have critical points where the derivative doesn't equal zero?
4. How do we find inflection points in a function?
5. How do critical points relate to concavity and inflection points?

Tip: Always check whether the derivative is undefined at certain points, as they may also be critical points.