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

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

❌

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

Переглянути це питання

Crowdly

Matematika - LS 24/25

Хочете миттєвий доступ до всіх перевірених відповідей на moodle.czu.cz?