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• Total sequences: \(2^9 = 512\)
• Let \(b_n\) be the number of binary sequences of length \(n\) with no two consecutive C’s (C = communal).
• Same recurrence: \(b_n = b_{n-1} + b_{n-2}\)
• \(b_1 = 2\) (C, I)
• \(b_2 = 3\) (II, IC, CI — not CC)
• \(b_3 = b_2 + b_1 = 3 + 2 = 5\)

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\log_2(x(x-4)) = 3 \Rightarrow x(x-4) = 2^3 = 8x^2 - 4x - 8 = 0Using the quadratic formula:x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 1 \times (-8)}}{2 \times 1}x = \frac{4 \pm \sqrt{16 + 32}}{2} = \frac{4 \pm \sqrt{48}}{2} = \frac{4 \pm 4\sqrt{3}}{2}x = 2 \pm 2\sqrt{3}Since \(x > 4\), \(x = 2 + 2\sqrt{3} \approx 5.464\).#### #### 5.464The function \( f(x) = x^3 - 6x^2 + 9x + c \) has a local maximum at \( x = 1 \). Find \( c \) if \( f(1) = 2 \).Solution 8: