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Určete Taylorův polynom druhého stupně funkce

$f(x)=\cos(7x-\pi)-9.$ f(x)=\cos(7x-\pi)-9. se středem v bodě $a=\frac{\pi}{6}$ a=\frac{\pi}{6}.

$-9+\frac{\sqrt 3}{2}-\frac{7}{2}(x-\frac{\pi}{6})-\frac{49\sqrt 3}{4}(x-\frac{\pi}{6})^2$ -9+\frac{\sqrt 3}{2}-\frac{7}{2}(x-\frac{\pi}{6})-\frac{49\sqrt 3}{4}(x-\frac{\pi}{6})^2

✅

$-9+\frac{\sqrt 3}{2}-\frac{7}{2}(x-\frac{\pi}{6})-\frac{49\sqrt 3}{2}(x-\frac{\pi}{6})^2$ -9+\frac{\sqrt 3}{2}-\frac{7}{2}(x-\frac{\pi}{6})-\frac{49\sqrt 3}{2}(x-\frac{\pi}{6})^2

❌

$-9+\frac{\sqrt 3}{2}-\frac{7}{2}(x+\frac{\pi}{6})+\frac{49\sqrt 3}{2}(x+\frac{\pi}{6})^2$ -9+\frac{\sqrt 3}{2}-\frac{7}{2}(x+\frac{\pi}{6})+\frac{49\sqrt 3}{2}(x+\frac{\pi}{6})^2

❌

$-9+\frac{\sqrt 3}{2}-\frac{1}{2}(x-\frac{\pi}{6})+\frac{\sqrt 3}{4}(x-\frac{\pi}{6})^2$ -9+\frac{\sqrt 3}{2}-\frac{1}{2}(x-\frac{\pi}{6})+\frac{\sqrt 3}{4}(x-\frac{\pi}{6})^2

❌

$-9+\frac{\sqrt 3}{2}-\frac{7}{2}(x-\frac{\pi}{6})+\frac{49\sqrt 3}{2}(x-\frac{\pi}{6})^2$ -9+\frac{\sqrt 3}{2}-\frac{7}{2}(x-\frac{\pi}{6})+\frac{49\sqrt 3}{2}(x-\frac{\pi}{6})^2

❌

$-9+\frac{\sqrt 3}{2}-\frac{7}{2}(x-\frac{\pi}{6})+\frac{49\sqrt 3}{4}(x-\frac{\pi}{6})^2$ -9+\frac{\sqrt 3}{2}-\frac{7}{2}(x-\frac{\pi}{6})+\frac{49\sqrt 3}{4}(x-\frac{\pi}{6})^2

❌

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Matematika - LS 24/25

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