Find domain of $ \sin ^ {-1} [\log_2(\frac{x}{2})]$

Question

Problem: Find domain of $ \sin ^ {-1} [\log_2(\frac{x}{2})]$

Solution: $\log_2(\frac{x}{2})$ is defined for $\frac{x}{2} > 0$

$\log_2(\frac{x}{2})$ is defined for $x > 0$

Also domain of $\sin ^ {-1}x$ is $[-1,1]$

When $x=1$ ,then $\log_2(\frac{x}{2})$ becomes $-1$

When $x=4$ ,then $\log_2(\frac{x}{2})$ becomes $1$

So domain is $[1,4]$

Check the graph. I think it is $\pi/2$ to positive infinity. — Torsten Hĕrculĕ Cärlemän, Jul 16 '13 at 13:04

score 6 · Accepted Answer · answered Jul 16 '13 at 13:09

6

Hint:

You can search the proper $x$ in which we can have $$\log_2(x/2)\in[-1,+1], ~~\text{and}~~~x/2>0$$ simultaneously. Note that the logarithm function is a one-one function here.

answered Jul 16 '13 at 13:09

Mikasa

67,374

1

It is also monotone increasing, which facilitates handling inequalities. – ncmathsadist Jul 16 '13 at 13:12
@rst: When $-1\le\log_2(x/2)\le 1 $ then according to comment below my answer you can write $2^{-1}\le x/2\le 2^1$ – Mikasa Jul 16 '13 at 13:20
1

Thanks, now it's more clear to me – rst Jul 16 '13 at 13:26

Suraj M S · Answer 2 · 2013-07-16T13:22:27.793

1

the domain is [1,4] as sin (y) is between -1 &1

edited Jul 16 '13 at 13:22

answered Jul 16 '13 at 13:13

Suraj M S

1,891

Can you explain in some more detail ? – rst Jul 16 '13 at 13:14

score 1 · Answer 3 · answered Jul 16 '13 at 13:20

1

here log(x/2) should be between -1&1 because if arc sin a=b then

sin (b)=a or a is between -1 &1

answered Jul 16 '13 at 13:20

Suraj M S

1,891

2

You could make your previous answer fixed. :) – Mikasa Jul 16 '13 at 13:21

Find domain of $ \sin ^ {-1} [\log_2(\frac{x}{2})]$

3 Answers3