Question

Find the average rate of change of f from 0 to (11pi)/12.
f(x)=tan(2x)

Answer 1

\[\frac{f(0)-f(\frac{11 \pi}{12})}{0-\frac{11 \pi}{12}}\]
it would have been nice if you didn't repost this twice so i wouldn't have to type that again

Answer 2

last time i did not include f(x)=tan(2x)

Answer 3

\[\frac{\tan(2 \cdot 0)-\tan(2 \cdot \frac{11 \pi}{12})}{-\frac{11 \pi}{12}}\]
\[\frac{0-\tan(\frac{11 \pi}{6})}{\frac{-11 \pi}{12}}\]
it is the same thing though
i mean it is the same formula to use

Answer 4

\[\frac{ \tan(\frac{11 \pi}{6})}{\frac{11 \pi}{12}} \cdot \frac{12}{12}=\frac{ 12 \tan(\frac{11 \pi}{6})}{11 \pi}\]

Answer 5

((-11pi)sqrt3)/36

Answer 6

\[\frac{12 \tan(\frac{6 \pi}{6}+\frac{5 \pi}{6})}{11 \pi}\]
\[\frac{12 \tan( \frac{5\pi}{6}+\pi)}{11 \pi}\]