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If [tex][tex]$a(x)=3x+1$[/tex][/tex] and [tex][tex]$b(x)=\sqrt{x-4}$[/tex][/tex], what is the domain of [tex][tex]$(b \circ a)(x)$[/tex][/tex]?

A. [tex]$(-\infty, \infty)$[/tex]
B. [tex]$[0, \infty)$[/tex]
C. [tex]$[1, \infty)$[/tex]
D. [tex]$[4, \infty)$[/tex]


Sagot :

To determine the domain of the composite function [tex]\( (b \circ a)(x) \)[/tex], where [tex]\( (b \circ a)(x) = b(a(x)) \)[/tex]:

1. Define the functions:
- [tex]\( a(x) = 3x + 1 \)[/tex]
- [tex]\( b(x) = \sqrt{x - 4} \)[/tex]

2. Understand the domain of [tex]\( b(x) \)[/tex]:
- For [tex]\( b(x) = \sqrt{x - 4} \)[/tex] to be defined, the expression inside the square root must be non-negative: [tex]\( x - 4 \geq 0 \)[/tex].
- Therefore, [tex]\( x \geq 4 \)[/tex].

3. Determine the input to [tex]\( b(x) \)[/tex] via [tex]\( a(x) \)[/tex]:
- Since we are evaluating [tex]\( b(a(x)) \)[/tex], we substitute [tex]\( a(x) \)[/tex] into [tex]\( b(x) \)[/tex].
- Therefore, [tex]\( b(a(x)) = b(3x+1) \)[/tex].

4. Establish conditions for [tex]\( b \circ a(x) \)[/tex] to be defined:
- The input to [tex]\( b \)[/tex], which is [tex]\( 3x + 1 \)[/tex], must satisfy the domain requirement for [tex]\( b \)[/tex].
- Hence, [tex]\( 3x + 1 \geq 4 \)[/tex].

5. Solve for [tex]\( x \)[/tex]:
- Simplify the inequality [tex]\( 3x + 1 \geq 4 \)[/tex]:
[tex]\[ 3x + 1 \geq 4 \][/tex]
[tex]\[ 3x \geq 3 \][/tex]
[tex]\[ x \geq 1 \][/tex]

Therefore, the domain of [tex]\( (b \circ a)(x) \)[/tex] is [tex]\( [1, \infty) \)[/tex].

Hence, the correct answer is:
[tex]\[ [1, \infty) \][/tex]