Matematika - LS 24/25 Course Materials & Test Answers | moodle.czu.cz

Určete Taylorův polynom druhého stupně funkce

$f(x)=\cos(5x-\pi)+8.$ f(x)=\cos(5x-\pi)+8. se středem v bodě $a=\frac{\pi}{3}$ a=\frac{\pi}{3}.

$\frac{15}{2}-\frac{\sqrt 3}{2}(x-\frac{\pi}{3})-\frac{1}{4}(x-\frac{\pi}{3})^2$ \frac{15}{2}-\frac{\sqrt 3}{2}(x-\frac{\pi}{3})-\frac{1}{4}(x-\frac{\pi}{3})^2

$\frac{15}{2}-\frac{5\sqrt 3}{2}(x-\frac{\pi}{3})+\frac{25}{4}(x-\frac{\pi}{3})^2$ \frac{15}{2}-\frac{5\sqrt 3}{2}(x-\frac{\pi}{3})+\frac{25}{4}(x-\frac{\pi}{3})^2

$\frac{15}{2}-\frac{5\sqrt 3}{2}(x-\frac{\pi}{3})-\frac{25}{2}(x-\frac{\pi}{3})^2$ \frac{15}{2}-\frac{5\sqrt 3}{2}(x-\frac{\pi}{3})-\frac{25}{2}(x-\frac{\pi}{3})^2

$\frac{15}{2}-\frac{5\sqrt 3}{2}(x-\frac{\pi}{3})+\frac{25}{2}(x-\frac{\pi}{3})^2$ \frac{15}{2}-\frac{5\sqrt 3}{2}(x-\frac{\pi}{3})+\frac{25}{2}(x-\frac{\pi}{3})^2

$\frac{15}{2}-\frac{5\sqrt 3}{2}(x-\frac{\pi}{3})-\frac{25}{4}(x-\frac{\pi}{3})^2$ \frac{15}{2}-\frac{5\sqrt 3}{2}(x-\frac{\pi}{3})-\frac{25}{4}(x-\frac{\pi}{3})^2

$\frac{15}{2}-\frac{5\sqrt 3}{2}(x+\frac{\pi}{3})-\frac{25}{2}(x+\frac{\pi}{3})^2$ \frac{15}{2}-\frac{5\sqrt 3}{2}(x+\frac{\pi}{3})-\frac{25}{2}(x+\frac{\pi}{3})^2

View this question

Crowdly

Matematika - LS 24/25

Want instant access to all verified answers on moodle.czu.cz?