Question

help please !!!!

explain using theorems 4 5 7 and 9 , why the fucntion is continous at every number in its domain , state the domain.


a(t) =arcsin(1+2t)

Answer 1

@Directrix

Answer 2

@mathmale

Answer 3

@dan815

Answer 4

who knows what the theorems are?
every function you know is continuous on its domain
the domain of arcsine is \[[-\frac{\pi}{2},\frac{\pi}{2}]\] set
\[1+2t=-\frac{\pi}{2}\] and solve for \(t\)
then do the same for \[\frac{\pi}{2}\] that will give you your domain

Answer 5

well not really im so lost in this lesson

Answer 6

why is it set to --pi/2?

Answer 7

i dont understand :(

Answer 8

because the domain of arcsine starts at \(-\frac{\pi}{2}\)

Answer 9

oh but i got -3pi/4

Answer 10

seems unlikely that if you solve \[1+2t=-\frac{\pi}{2}\] you get \[t=-\frac{3\pi}{4}\]

Answer 11

yes i got tht

Answer 12

you got what? \(-\frac{3\pi}{4}\)?

Answer 13

yes.. idk what i did wrong

Answer 14

takes two steps, just like solving \[1+2t=A\] for \(t\)
subtract 1, divide by 2

Answer 15

1/2

Answer 16

\[1+2t=A\\
2t=A-1\\t=\frac{A-1}{2}\]

Answer 17

replace \(A\) by \(-\frac{\pi}{2}\)

Answer 18

@satellite73 come back

Answer 19

@Directrix

Answer 20

What do these theorems state?

The theorem numbers are different in different texts.

>>theorems 4 5 7 and 9

Answer 21

well im not sure im so confuse