Матеріали курсу Matematika - LS 24/25 та відповіді на тести | moodle.czu.cz

Určete Taylorův polynom druhého stupně funkce

$f(x)=\cos(5x-\pi)+6.$ f(x)=\cos(5x-\pi)+6. se středem v bodě $a=\frac{\pi}{4}$ a=\frac{\pi}{4}.

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

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

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

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

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

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

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Matematika - LS 24/25

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