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Určete Taylorův polynom druhého stupně funkce

$f(x)=\sin(5x-\pi)+7.$ f(x)=\sin(5x-\pi)+7. se středem v bodě $a=\frac{\pi}{3}$ a=\frac{\pi}{3}.

$7+\frac{\sqrt 3}{2}-\frac{1}{2}(x-\frac{\pi}{3})-\frac{\sqrt 3}{4}(x-\frac{\pi}{3})^2$ 7+\frac{\sqrt 3}{2}-\frac{1}{2}(x-\frac{\pi}{3})-\frac{\sqrt 3}{4}(x-\frac{\pi}{3})^2

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$\frac{13}{2}-\frac{5\sqrt 3}{2}(x-\frac{\pi}{3})-\frac{25}{4}(x-\frac{\pi}{3})^2$ \frac{13}{2}-\frac{5\sqrt 3}{2}(x-\frac{\pi}{3})-\frac{25}{4}(x-\frac{\pi}{3})^2

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$7+\frac{\sqrt 3}{2}-\frac{5}{2}(x-\frac{\pi}{3})-\frac{25\sqrt 3}{2}(x-\frac{\pi}{3})^2$ 7+\frac{\sqrt 3}{2}-\frac{5}{2}(x-\frac{\pi}{3})-\frac{25\sqrt 3}{2}(x-\frac{\pi}{3})^2

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$7+\frac{\sqrt 3}{2}-\frac{5}{2}(x-\frac{\pi}{3})-\frac{25\sqrt 3}{4}(x-\frac{\pi}{3})^2$ 7+\frac{\sqrt 3}{2}-\frac{5}{2}(x-\frac{\pi}{3})-\frac{25\sqrt 3}{4}(x-\frac{\pi}{3})^2

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$7+\frac{\sqrt 3}{2}-\frac{5}{2}(x-\frac{\pi}{3})+\frac{25\sqrt 3}{4}(x-\frac{\pi}{3})^2$ 7+\frac{\sqrt 3}{2}-\frac{5}{2}(x-\frac{\pi}{3})+\frac{25\sqrt 3}{4}(x-\frac{\pi}{3})^2

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$7+\frac{\sqrt 3}{2}-\frac{5}{2}(x+\frac{\pi}{3})-\frac{25\sqrt 3}{2}(x+\frac{\pi}{3})^2$ 7+\frac{\sqrt 3}{2}-\frac{5}{2}(x+\frac{\pi}{3})-\frac{25\sqrt 3}{2}(x+\frac{\pi}{3})^2

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Matematika - LS 24/25

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