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Určete Taylorův polynom druhého stupně funkce

$f(x)=\cos(3x+\frac{\pi}{6})-7.$ f(x)=\cos(3x+\frac{\pi}{6})-7. se středem v bodě $a=\frac{\pi}{6}$ a=\frac{\pi}{6}.

$\frac{-15}{2}-\frac{3\sqrt 3}{2}(x-\frac{\pi}{6})+\frac{9}{2}(x-\frac{\pi}{6})^2$ \frac{-15}{2}-\frac{3\sqrt 3}{2}(x-\frac{\pi}{6})+\frac{9}{2}(x-\frac{\pi}{6})^2

$\frac{-15}{2}-\frac{3\sqrt 3}{2}(x-\frac{\pi}{6})-\frac{9}{2}(x-\frac{\pi}{6})^2$ \frac{-15}{2}-\frac{3\sqrt 3}{2}(x-\frac{\pi}{6})-\frac{9}{2}(x-\frac{\pi}{6})^2

$\frac{-15}{2}-\frac{3\sqrt 3}{2}(x-\frac{\pi}{6})+\frac{9}{4}(x-\frac{\pi}{6})^2$ \frac{-15}{2}-\frac{3\sqrt 3}{2}(x-\frac{\pi}{6})+\frac{9}{4}(x-\frac{\pi}{6})^2

$\frac{-15}{2}-\frac{3\sqrt 3}{2}(x-\frac{\pi}{6})-\frac{9}{4}(x-\frac{\pi}{6})^2$ \frac{-15}{2}-\frac{3\sqrt 3}{2}(x-\frac{\pi}{6})-\frac{9}{4}(x-\frac{\pi}{6})^2

$\frac{-15}{2}-\frac{\sqrt 3}{2}(x-\frac{\pi}{6})-\frac{1}{4}(x-\frac{\pi}{6})^2$ \frac{-15}{2}-\frac{\sqrt 3}{2}(x-\frac{\pi}{6})-\frac{1}{4}(x-\frac{\pi}{6})^2

$\frac{-15}{2}-\frac{3\sqrt 3}{2}(x+\frac{\pi}{6})-\frac{9}{2}(x+\frac{\pi}{6})^2$ \frac{-15}{2}-\frac{3\sqrt 3}{2}(x+\frac{\pi}{6})-\frac{9}{2}(x+\frac{\pi}{6})^2

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Matematika - LS 24/25

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