Question

state the domain and range of the function f(x)-arc3sin(x/2)

Answer 1

Could you write it in equation writer please?

Answer 2

\[f(x) = 3\sin^{-1} (x/2)\]

Answer 3

domain of arcsine is from [-1,1]

Answer 4

in your case you should have
\[-1\leq \frac{x}{2}\leq 1\] so
\[-2\leq x \leq 2\] is your domain

Answer 5

could you explain how you got the domain? the answers at the back said that but i dont quite get it

Answer 6

range of arcsine is
\[[-\frac{\pi}{2},\frac{\pi}{2}]\] so your range is
\[[-\frac{3\pi}{2},\frac{3\pi}{2}]\]

Answer 7

sure

Answer 8

if you want to take the arcsine of some number, that number has to be bigger than or equal to -1 and less than or equal to 1
this is clear yes?

Answer 9

Arcsine is defined for:
\[\forall x \in [-1;1], \arcsin(\sin(\theta))=\theta \]
\[\rightarrow -1 \le \frac{x}{2} \le 1 \iff -2 \le x \le 2\]
Domain is therefore
\[D=[-2;2]\]

Answer 10

but this function says first divide by 2. it is
\[3\sin^{-1}(\frac{x}{2})\] sp of you divide by 2 first you are allowed numbers that when divided by 2 are between -1 and .
therefore you numbers (x itself) has to be between -2 and 2

Answer 11

some typos in what i wrote, but i hope it is clear

Answer 12

thank you, both of you guys. i'm still trying to digest it, but ill get it eventually.

Answer 13

? What do you mean it says it first?
x/2 is the value of a sine, and the value of a sine HAS to be in [-1;1]... yet
\[x \in [-2;2] \iff \frac{x}{2} \in [-1;1]\]
Which is why the domain is [-2;2]

Answer 14

someone1348, I actually understood that. thanks guys

Answer 15

hope it is clear what i mean. how to i evaluate
\[3\sin^{-1}(\frac{x}{2})\] for a given number replacement for x?
first divide it by 2
then take the inverse sine
then multiply by 3.

so "first divide by 2" yes?

Answer 16

\[-1<sinx<1\]
the range of sinx is [-1,1]
so the domain of sine inverse of x is [-1,1]

Answer 17

can you also explain how to get the range?

Answer 18

The range is the interval from the lowest value to the highest value.
The lowest value arcsine can take is -pi/2, the highest is pi/2

It's the image interval by the function of the domain interval.

Answer 19

ok sinx is not 1 to 1
so we want sine inverse of x to exist
so we need to restrict the domain of sine so that it is 1 to 1 so that the inverse does exist
the restricted domain of sinx is [-pi/2,pi/2]

Answer 20

so the range of sine inverse is [-pi/2,pi/2]

Answer 21

In other words,
\[\arcsin([-1;1])=[-\frac{\pi}{2};\frac{\pi}{2}]\]

Answer 22

i know how it works for just sinx, but what happens when you have a 3 in front of it

Answer 23

like, for f(x)=inverse 3sin(x/2) in particular

Answer 24

right! that is why you have to multiply by 3 to get your range. here is a nice picture (ignore the second one) and you will see what the range is
http://www.wolframalpha.com/input/?i=y%3D3+arcsin%28x%2F2%29

Answer 25

Well then it all gets multiplied by 3.. I mean, if the range of arcsin is
\[[-\frac{\pi}{2};\frac{\pi}{2}]\]
Then the range of 3 times arcsin is
\[[-\frac{3 \pi}{2};\frac{3\pi}{2}]\]

Answer 26

OHHHHHHH I get it, thank you guys, really.

Answer 27

since the range of arcsine is
\[[-\frac{\pi}{2},\frac{\pi}{2}]\] the range of 3 arcsine is
\[[-\frac{3\pi}{2},\frac{3\pi}{2}]\]