How do you find the domain for (foh)(x)= sqrt4-(x)^2 ?

Question

anonymous · Answer

does the square root goo all the way through the equation?

anonymous · Answer

Yes

anonymous · Answer

alright so this makes it much easier. 
Since your square root goes through the whole equation, it means the value inside the square root cannot be negative. Given the equation $$f(x) = \sqrt{4 - x^2}$$

anonymous · Answer

x has to be a value between -2 and 2, since it cannot result in a value greater than 4 after doing the exponential operation. Therefore -2<x<2 is your domain.

anonymous · Answer

Thanks ! Would i do the same thing for $$\sqrt{(4-x)^2}$$

anonymous · Answer

No that is a set of all real numbers. Because even if x is greater than 4, squaring the negative would result in a positive value which can be square rooted. 

Say for instance x = 5 

$$f(x) = \sqrt{(4 - x)^2}$$
$$f(5) = \sqrt{(4-5)^2}$$
$$f(5) = \sqrt{(4 -5)^2}$$
$$f(5) = \sqrt{(-1)^2}$$
$$f(5) = \sqrt{(1)}$$
\[f(5) = 1

anonymous · Answer

$$f(5)$$
********

anonymous · Answer

Oh okay thank you :D!

anonymous · Answer

no problem and nice pic :) :P

anonymous · Answer

Thanks :)

anonymous · Answer

I have one last question.. lol

anonymous · Answer

lol sure ask away :)

anonymous · Answer

Okay thanks lol 

Given: f(x)=$$\sqrt{4-x} $$ and h(x)= x^2 

Find g, the inverse of f. GIve the domain and range for both f and g, sketch the graphs of f and g on the same set of axes. 


Even if you could just help me with the domain and range that'd be awesome lol

anonymous · Answer

IM SOO SORRY, i got carried away by my own hmwk

anonymous · Answer

i havent got to my inverse hmwk yet, so i may not be the best source for the answer. :( 
Sorry once again :/