Functions and Domains

[neversummer711]

Been learning domains and functions needless to say im stumped on how to write the answer to this problem

(G/h)(x)

g(x) = x^2-9
h(x) = x^2-16

so itd be x^2-9
x^2-16

so to find the domain of the function it'd be x^2-16<0 ? which then would be x^2>16, then x>4 How do I write this? as the domain.... Would it be D: [4,Infinity) ?

 

[pka]

(G/h)(x)
g(x) = x^2-9
h(x) = x^2-16

I am not sure what this question is about.
If \(\displaystyle F(x)=\dfrac{x^2-9}{x^2-16}\) then the only difficulty there is possibility of division by zero.

So the domain is \(\displaystyle \mathbb{R}\setminus\{4,-4\}=(-\infty,-4)\cup(-4,4)\cup(4,\infty)~.\)

 

[neversummer711]

awesome! I finally figured it out and that was def what I got

Im really now getting stumped on composite functions

ex... (jog)(x)

j(x) = √(x+5)
g(x) = x²-9

so do I set it up like j[g(x)] so it'd look like √[(x²-9)+5?

 

[pka]

Im really now getting stumped on composite functions
ex... (jog)(x)
j(x) = √(x+5)
g(x) = x²-9
so do I set it up like j[g(x)] so it'd look like √[(x²-9)+5?

What is the domain of \(\displaystyle \sqrt{x^2-4}~?\)

 

[neversummer711]

So then it would be √x²-4 would be (x+2)(x-2). Would make the domain (-∞,-2)u(2,-2)u(2,∞)?[FONT=MathJax_Main][/FONT]

 

[pka]

So then it would be √x²-4 would be (x+2)(x-2). Would make the domain (-∞,-2)u(2,-2)u(2,∞)?

NO! \(\displaystyle x=0\) makes it undefined. \(\displaystyle \sqrt{0^2-4}\) is not defined.

So redo it.

 

[neversummer711]

[FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Main][/FONT]

 

[neversummer711]

√x²-4

x²-4<0
x²>4
x>√4
x>2?