https://file.aiquickdraw.com/mathbot/file/17276549412368jook7x7.jpg

The function provided in the image is:

\[
f(x) = 3x^2 - 8x
\]

To find the critical numbers of the function, we follow these steps:

1. **Find the derivative of the function:**  
   \[
   f'(x) = \frac{d}{dx}(3x^2 - 8x)
   \]
   Using the power rule:
   \[
   f'(x) = 6x - 8
   \]

2. **Set the derivative equal to zero to find critical points:**
   \[
   6x - 8 = 0
   \]
   Solving for \(x\):
   \[
   6x = 8
   \]
   \[
   x = \frac{8}{6} = \frac{4}{3}
   \]

3. **Verify that this is a critical point:**  
   Critical points occur where the derivative is zero or undefined. In this case, \(f'(x) = 0\) at \(x = \frac{4}{3}\), and the derivative is defined for all real numbers, so \(x = \frac{4}{3}\) is the only critical number.

Thus, the critical number is:
\[
x = \frac{4}{3}
\]

Would you like more details on any of these steps or further clarification?

Here are 5 related questions to deepen your understanding:
1. What are critical points, and why are they important in calculus?
2. How do you determine if a critical point is a maximum, minimum, or saddle point?
3. Can critical points exist where the derivative is undefined?
4. What is the difference between absolute and relative extrema?
5. How do higher-order derivatives help in classifying critical points?

**Tip:** Always check the second derivative to determine whether a critical point is a local minimum or maximum.

Find the critical numbers of the function f(x) = 3x^2 - 8x. (Enter your answers as a comma-separated list.)

Learn how to find the critical numbers for the quadratic function f(x) = 3x^2 - 8x using the derivative and solving for where f'(x) = 0. Suitable for high school and early college students learning calculus.

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