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

Určete Taylorův polynom druhého stupně funkce

$f(x)=\cos(3x-\frac{\pi}{3})-1.$ f(x)=\cos(3x-\frac{\pi}{3})-1. se středem v bodě $a=\frac{\pi}{6}$ a=\frac{\pi}{6}.

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

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

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

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

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

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

View this question

Crowdly

Matematika - LS 24/25

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