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Určete Taylorův polynom druhého stupně funkce

$f(x)=\cos(3x+\frac{\pi}{2})+2.$ f(x)=\cos(3x+\frac{\pi}{2})+2. se středem v bodě $a=-\frac{\pi}{4}$ a=-\frac{\pi}{4}.

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

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$2+\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2}(x+\frac{\pi}{4})+\frac{\sqrt 2}{4}(x+\frac{\pi}{4})^2$ 2+\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2}(x+\frac{\pi}{4})+\frac{\sqrt 2}{4}(x+\frac{\pi}{4})^2

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

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

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

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

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Matematika - LS 24/25

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