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Určete Taylorův polynom druhého stupně funkce

$f(x)=\cos(3x-\frac{\pi}{3})-8.$ f(x)=\cos(3x-\frac{\pi}{3})-8. se středem v bodě $a=\frac{\pi}{6}$ a=\frac{\pi}{6}.

$-8+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{6})-\frac{9\sqrt 3}{2}(x-\frac{\pi}{6})^2$ -8+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{6})-\frac{9\sqrt 3}{2}(x-\frac{\pi}{6})^2

$-8+\frac{\sqrt 3}{2}-\frac{3}{2}(x+\frac{\pi}{6})+\frac{9\sqrt 3}{2}(x+\frac{\pi}{6})^2$ -8+\frac{\sqrt 3}{2}-\frac{3}{2}(x+\frac{\pi}{6})+\frac{9\sqrt 3}{2}(x+\frac{\pi}{6})^2

$-8+\frac{\sqrt 3}{2}-\frac{1}{2}(x-\frac{\pi}{6})+\frac{\sqrt 3}{4}(x-\frac{\pi}{6})^2$ -8+\frac{\sqrt 3}{2}-\frac{1}{2}(x-\frac{\pi}{6})+\frac{\sqrt 3}{4}(x-\frac{\pi}{6})^2

$-8+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{6})-\frac{9\sqrt 3}{4}(x-\frac{\pi}{6})^2$ -8+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{6})-\frac{9\sqrt 3}{4}(x-\frac{\pi}{6})^2

$-8+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{6})+\frac{9\sqrt 3}{4}(x-\frac{\pi}{6})^2$ -8+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{6})+\frac{9\sqrt 3}{4}(x-\frac{\pi}{6})^2

$-8+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{6})+\frac{9\sqrt 3}{2}(x-\frac{\pi}{6})^2$ -8+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{6})+\frac{9\sqrt 3}{2}(x-\frac{\pi}{6})^2

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Matematika - LS 24/25

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