Question

the function $f(x) = 2x^3 - 33x^2 + 108x - 1$ has two critical numbers.
the smaller one is $x = \square$
and the larger one is $x = \square$
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Explanation:

Step1: Find the derivative of \( f(x) \)

To find critical numbers, we first find the derivative of \( f(x) = 2x^3 - 33x^2 + 108x - 1 \). Using the power rule, the derivative \( f'(x) \) is:
\( f'(x) = 6x^2 - 66x + 108 \)

Step2: Simplify the derivative

We can factor out a 6 from the derivative:
\( f'(x) = 6(x^2 - 11x + 18) \)

Step3: Factor the quadratic

Now, factor the quadratic \( x^2 - 11x + 18 \). We need two numbers that multiply to 18 and add to -11. Those numbers are -2 and -9. So:
\( x^2 - 11x + 18 = (x - 2)(x - 9) \)
Thus, \( f'(x) = 6(x - 2)(x - 9) \)

Step4: Find critical numbers

Critical numbers occur where \( f'(x) = 0 \) or \( f'(x) \) is undefined. Since \( f'(x) \) is a polynomial, it's defined everywhere. Set \( f'(x) = 0 \):
\( 6(x - 2)(x - 9) = 0 \)
Dividing both sides by 6 (which doesn't change the solution), we get:
\( (x - 2)(x - 9) = 0 \)
Setting each factor equal to zero gives \( x - 2 = 0 \) or \( x - 9 = 0 \), so \( x = 2 \) or \( x = 9 \)

Answer:

The smaller one is \( x = 2 \) and the larger one is \( x = 9 \)