Question

Find domain of f/g

Answer 1

\[f(x)=\sqrt{4-x^2}\]
\[g(x)=\sqrt{3x+4}\]

Answer 2

f/g = (x² - 1)/(2x + 3)

The domain are the values of x that are permitted in this ratio. Those would be the values of 'x' that do not give zeros in the denom.

This happens at: 2x + 3 = 0

==> 2x = -3
==> x = -3/2

the domain 'x' excluding -3/2. In proper terms the domain are all real numbers except x = -3/2.

Answer 3

'-'

Answer 4

I understand that but I don't know what to do when there's square roots

Answer 5

ohhhh

Answer 6

http://www.purplemath.com/modules/radicals.htm <-- that might help you :)

Answer 7

$$\Huge \frac{a^m}{b^m}=(\frac{a}{b})^m$$

Answer 8

Can someone walk me through the problem I posted?

Answer 9

i kinda did but ill just let someone else do it cuz i guess i didnt make sense '-'

Answer 10

well because g(x) is a square root function, its argument must be at least 0, and because g(x) is in the denominator, g(x) can not be 0, so its argument now is strictly positive.

f(x) is also a square root function, so its argument is at least 0.

so you have,
{ x | 4-x^2 ≥ 0 AND 3x + 4 > 0 }

Answer 11

@sourwing 's reply is incomplete, for your information. Try to work on
\[{ x \in \mathbb{R} | 4-x^2 \geq 0,~ 3x + 4 > 0 }\]