You've done a thorough analysis of the function \( f(x) = x^3 - 9x^2 + 24x - 1 \) over the interval \([3, 6]\). Let's summarize and confirm the results step by step:
Step 1: Evaluate at the Endpoints
At \( x = 3 \): \[ f(3) = 3^3 - 9 \cdot 3^2 + 24 \cdot 3 - 1 = 27 - 81 + 72 - 1 = 17 \]
At \( x = 6 \): \[ f(6) = 6^3 - 9 \cdot 6^2 + 24 \cdot 6 - 1 = 216 - 324 + 144 - 1 = 35 \]
Step 2: Find the Critical Points
- Calculate the derivative: \[ f'(x) = 3x^2 - 18x + 24 \] Setting the derivative to zero: \[ 3x^2 - 18x + 24 = 0 \] Dividing by 3: \[ x^2 - 6x + 8 = 0 \] Factoring the quadratic: \[ (x - 2)(x - 4) = 0 \] Therefore, \( x = 2 \) and \( x = 4 \).
Step 3: Evaluate at Critical Points
Only \( x = 4 \) lies within the interval \([3, 6]\):
\[ f(4) = 4^3 - 9 \cdot 4^2 + 24 \cdot 4 - 1 = 64 - 144 + 96 - 1 = 15 \]
Step 4: Compare Values
Now, we compare the function values at the endpoints and the critical point:
- \( f(3) = 17 \)
- \( f(4) = 15 \)
- \( f(6) = 35 \)
Conclusion
- Maximum value: \( 35 \) at \( x = 6 \)
- Minimum value: \( 15 \) at \( x = 4 \)
Thus, the maximum and minimum values of the function on the interval \([3, 6]\) are indeed:
- Maximum: \( 35 \) (at \( x = 6 \))
- Minimum: \( 15 \) (at \( x = 4 \))
Your analysis is accurate and well-structured!