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Určete Taylorův polynom druhého stupně funkce

$f(x)=\cos(5x+\frac{\pi}{2})-2.$ f(x)=\cos(5x+\frac{\pi}{2})-2. se středem v bodě $a=-\frac{\pi}{6}$ a=-\frac{\pi}{6}.

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

$\frac{-3}{2}+\frac{5\sqrt 3}{2}(x+\frac{\pi}{6})+\frac{25}{2}(x+\frac{\pi}{6})^2$ \frac{-3}{2}+\frac{5\sqrt 3}{2}(x+\frac{\pi}{6})+\frac{25}{2}(x+\frac{\pi}{6})^2

$\frac{-3}{2}+\frac{5\sqrt 3}{2}(x+\frac{\pi}{6})+\frac{25}{4}(x+\frac{\pi}{6})^2$ \frac{-3}{2}+\frac{5\sqrt 3}{2}(x+\frac{\pi}{6})+\frac{25}{4}(x+\frac{\pi}{6})^2

$\frac{-3}{2}+\frac{5\sqrt 3}{2}(x-\frac{\pi}{6})+\frac{25}{2}(x-\frac{\pi}{6})^2$ \frac{-3}{2}+\frac{5\sqrt 3}{2}(x-\frac{\pi}{6})+\frac{25}{2}(x-\frac{\pi}{6})^2

$\frac{-3}{2}+\frac{5\sqrt 3}{2}(x+\frac{\pi}{6})-\frac{25}{4}(x+\frac{\pi}{6})^2$ \frac{-3}{2}+\frac{5\sqrt 3}{2}(x+\frac{\pi}{6})-\frac{25}{4}(x+\frac{\pi}{6})^2

$\frac{-3}{2}+\frac{5\sqrt 3}{2}(x+\frac{\pi}{6})-\frac{25}{2}(x+\frac{\pi}{6})^2$ \frac{-3}{2}+\frac{5\sqrt 3}{2}(x+\frac{\pi}{6})-\frac{25}{2}(x+\frac{\pi}{6})^2

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Matematika - LS 24/25

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