Respuesta :
a(x) = 3x + 1
b(x) = √(x-4)
Therefore
[tex](b \circ a) (x) = \sqrt{3x+1-4} =\sqrt{3x-3} [/tex]
This composition of functions is only defined as a real number when x≥0.
Therefore its domain is x = [1,∞]
Answer: [1,∞]
b(x) = √(x-4)
Therefore
[tex](b \circ a) (x) = \sqrt{3x+1-4} =\sqrt{3x-3} [/tex]
This composition of functions is only defined as a real number when x≥0.
Therefore its domain is x = [1,∞]
Answer: [1,∞]
Answer: The correct option is C. The domain of [tex](b\circ a)(x)[/tex] is [tex][0,\infty)[/tex].
Explanation:
It is given that,
[tex]a(x)=3x+1[/tex]
[tex]b(x)=\sqrt{x-4}[/tex]
First we have to find the composite function [tex](b\circ a)(x)[/tex].
[tex](b\circ a)(x)=b(a(x))[/tex]
[tex](b\circ a)(x)=b(3x+1)[/tex]
[tex](b\circ a)(x)=\sqrt{3x+1-4}[/tex]
[tex](b\circ a)(x)=\sqrt{3x-3}[/tex]
We know that a root function is defined for only positive values therefore the composite function is defined if,
[tex]3x-3\geq 0[/tex]
[tex]3x\geq 3[/tex]
[tex]x\geq 1[/tex]
The function is defined for all the values of x which are greater than or equal to 1. Therefore the domain of [tex](b\circ a)(x)[/tex] is [tex][0,\infty)[/tex] and option C is correct.