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📅 March 9, 2026 👤 scraface

Mar 09, 2026

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Since \(x > 4\), \(x = 2 + 2\sqrt{3} \approx 5.464\).
#### #### 5.464
The function \( f(x) = x^3 - 6x^2 + 9x + c \) has a local maximum at \...
Solution 8:
\( f'(x) = 3x^2 - 12x + 9 \).
Setting \( f'(x) = 0 \) for critical points:
3x^2 - 12x + 9 = 0 \Rightarrow x^2 - 4x + 3 = 0 \Rightarrow (x-3)(x-1)...
So \( x = 1 \) and \( x = 3 \) are critical points.
Check \( f(1) = 2 \):
f(1) = 1^3 - 6(1)^2 + 9(1) + c = 1 - 6 + 9 + c = 4 + c

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