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To find the critical numbers of the given function:

\[
f(x) = 2x^3 - 24x^2 + 42x - 3
\]

### Step 1: Find the derivative of \( f(x) \)
Critical points occur where the derivative is zero or undefined. First, we calculate the derivative of \( f(x) \):

\[
f'(x) = \frac{d}{dx} (2x^3 - 24x^2 + 42x - 3)
\]
Using the power rule:

\[
f'(x) = 6x^2 - 48x + 42
\]

### Step 2: Solve \( f'(x) = 0 \)
Now, set the derivative equal to zero to find the critical points:

\[
6x^2 - 48x + 42 = 0
\]

### Step 3: Simplify and solve the quadratic equation
We can divide the entire equation by 6 to simplify:

\[
x^2 - 8x + 7 = 0
\]

Now, solve this quadratic equation using the quadratic formula:

\[
x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(7)}}{2(1)}
\]
\[
x = \frac{8 \pm \sqrt{64 - 28}}{2}
\]
\[
x = \frac{8 \pm \sqrt{36}}{2}
\]
\[
x = \frac{8 \pm 6}{2}
\]

So the two solutions are:

\[
x = \frac{8 + 6}{2} = 7 \quad \text{and} \quad x = \frac{8 - 6}{2} = 1
\]

### Step 4: Conclusion
The smaller critical number is \( x = 1 \) and the larger critical number is \( x = 7 \).

Would you like further clarification on any step, or additional details?

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Here are some related questions:
1. What is the significance of critical points in calculus?
2. How can you classify critical points as minima, maxima, or saddle points?
3. What role does the second derivative play in analyzing critical points?
4. How does one use critical points in optimization problems?
5. What happens if a function's derivative is undefined at a critical point?

**Tip:** Always check the second derivative to determine whether the critical points are maxima, minima, or points of inflection!

The function f(x) = 2x^3 - 24x^2 + 42x - 3 has two critical numbers. The smaller one is x = and the larger one is x =.

Learn how to find the critical numbers of the cubic function f(x) = 2x^3 - 24x^2 + 42x - 3 using derivatives and the quadratic formula. A step-by-step guide for high school calculus students (Grades 10-12).

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