The sequence is "xxx333xx". The third digit cannot be 3, so it must be 4 (1 choice). The seventh digit must be 4, and the last digit can be either 3 or 4, giving \(1 \times 2 = 2\) configurations.
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- If "333" starts at position 3:
- The sequence is "xx333xxx". The second digit cannot be 3, so it must be 4 (1 choice). The sixth digit must be 4, and the last 2 digits can be either 3 or 4, giving \(1 \times 2^2 = 4\) configurations.
- If "333" starts at position 4:
- If "333" starts at position 5:
- The sequence is "xxxx333x". The fourth digit must be 4 (1 choice), and the eighth digit can be either 3 or 4 (2 choices). The first 3 digits can be either 3 or 4, but must not form "333", which leaves \(2^3 - 1 = 7\) configurations
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Define $ g(u) = 3u^2 - 4u^4 $, $ u \in [-1, 1] $.
g'(u) = 6u - 16u^3 = 2u(3 - 8u^2)
Set $ g'(u) = 0 $: $ u = 0 $ or $ 3 - 8u^2 = 0 \Rightarrow u^2 = rac{3}{8} \Rightarrow u = \pm \sqrt{ rac{3}{8}} = \pm rac{\sqrt{6}}{4} $
Evaluate $ g(u) $:
At $ u = 0 $: $ g = 0 $
At $ u = \pm rac{\sqrt{6}}{4} $:
g = 3\left( rac{6}{16}
ight) - 4\left( rac{36}{256}
ight) = rac{18}{16} - rac{144}{256} = rac{9}{8} - rac{9}{16} = rac{18 - 9}{16} = rac{9}{16} = 0.5625
So maximum of $ \sin 3x \sin x $ is $ rac{9}{16} $, thus minimum of $ f(x) = 1 - rac{9}{16} = rac{7}{16} $? But wait â this contradicts earlier thought.