[tex]\boxed{ \ x \geq 1 \ }[/tex] or can be written as [tex]\boxed{ \ [1, \infty) \ }[/tex]
This is a question about the composition of functions and how to get a domain function.
Given [tex]\boxed{ \ a(x) = 3x + 1 \ }[/tex] and [tex]\boxed{ \ b(x) = \sqrt{x - 4} \ }[/tex].
We will form (b o a)(x) and then determine the domain.
Step-1
[tex]\boxed{ \ (b \circ a)(x) = b(a(x)) \ }[/tex]
Replace each appearance of x in b(x) with [tex]\boxed{ \ a(x) = 3x + 1 \ }[/tex].
[tex]\boxed{ \ (b \circ a)(x) = \sqrt{(3x + 1) - 4} \ }[/tex]
Thus, [tex]\boxed{ \ (b \circ a)(x) = \sqrt{3x - 3} \ }[/tex]
Step-2
To be defined, the value under the radical sign must not be negative. Therefore, the domain of [tex](b \circ a)(x) = \sqrt{3x - 3}[/tex] are processed as follows.
[tex]\boxed{ \ 3x - 3 \geq 0 \ }[/tex]
Both sides added by 3.
[tex]\boxed{ \ 3x \geq 3 \ }[/tex]
Both sides divided by 3.
[tex]\boxed{ \ x\geq 1 \ }[/tex]
Thus, the domain of [tex](b \circ a)(x) = \sqrt{3x - 3}[/tex] is [tex]\boxed{ \ x \geq 1 \ }[/tex] or can be written as [tex]\boxed{ \ [1, \infty) \ }[/tex]
Keywords: composition of function, if a(x) = 3x + 1, and, b(x) = √(x-4), what is the domain of, (b o a)(x), b(a(x)), defined, the value, under the radical sign, must not be negative,
