# Math Help - Find the domain - question

1. ## Find the domain - question

I know it's a dumb question, but could you please show me the right logic behind those calculations in order to find the domains of the following composite questions - I found the answers, but eventually making a lot of mistakes, so I am not sure in how I solved them; I just want to make clear once and forever on those points. Thanks

f(x) = sqrt(3x-8)
D = [8/3, INF)

and
g(x) = 5x^2 -1
D= R

1. (f o g)(x) = f(g(x)) =f(5x^2 - 1) = sqrt(3(5x^2 - 1) - 8) = 15x^2 – 3 – 8 = sqrt(15x^2 -11)

D= (-INF, -sqrt(11/15)]U[sqrt(11/15), INF)

2. (g o f)(x) = g(f(x)) = 5(sqrt(3x - 8))^2 -1 = 15x -40 – 1 = 15x-41;

D=[8/3, inf)

3. (f o f)(x) = sqrt(3(sqrt(3x - 8)) - 8)

D=[136/27, INF)

2. ## Re: Find the domain - question

The "default" domain of a given "formula" is the largest set of numbers for which it can be calculted.

In order to be in the domain of a composition of functions, fog(x)= f(gx)), x must be in the domain of g and y= g(x) must be in the domain of f.
Here, $g(x)= 5^2- 1$ and we can certainly square any real number, and then subtract 1. Any number is in the domain of f. $f(x)= \sqrt{3x- 8}$ which has domain all x such that $3x- 8\ge 0$. Okay, replacing that "x" with g(x) we are saying that $3g(x)- 8= 3(5x^2- 1)- 8= 15x^2- 11\ge 0$. Can you solve $15x^2\ge 11$?

g o f(x) is comparatively easy- q has domain all real numbers so the domain of the composition is just the domain of f.

f o f(x) has domain is all x such that 3x- 8\ge 0 and $\sqrt{3x- 8}\ge 0$.

3. ## Re: Find the domain - question

[QUOTE=HallsofIvy;779894]The "default" domain of a given "formula" is the largest set of numbers for which it can be calculted.

In order to be in the domain of a composition of functions, fog(x)= f(gx)), x must be in the domain of g and y= g(x) must be in the domain of f.
Here, $g(x)= 5^2- 1$ and we can certainly square any real number, and then subtract 1. Any number is in the domain of f. $f(x)= \sqrt{3x- 8}$ which has domain all x such that $3x- 8\ge 0$. Okay, replacing that "x" with g(x) we are saying that $3g(x)- 8= 3(5x^2- 1)- 8= 15x^2- 11\ge 0$. Can you solve $15x^2\ge 11$?

g o f(x) is comparatively easy- q has domain all real numbers so the domain of the composition is just the domain of f.

f o f(x) has domain is all x such that $3x- 8\ge 0 and sqrt{3x- 8}$ [QUOTE]

yes, $15x^2\ge 11 = x^2\ge {11}/{15} = x\ge \ pos/neg\sqrt{{11}/{15}}$ , I got it , it's more clear now

and with the third one $3x- 8\ge 0, \sqrt{3x- 8}\ge 0$ - I'm more confused with that one