Answer:
[tex][1,\infty)[/tex]
Step-by-step explanation:
[tex]b(x)=\sqrt{x-4}[/tex]
[tex]a(x)=3x+1[/tex]
Since we want to know the domain of [tex](b \circ a)(x)[/tex], let's first consider the domain of the inside function, that is, that of [tex]a(x)=3x+1[/tex]. Every polynomial function has domain all real numbers.
So we can plug anything for function [tex]a[/tex] and get a number back.
Now the other function is going to be worrisome because it has a square root. You cannot take square root of negative numbers if you are only considering real numbers which that is the case with most texts.
Let's find [tex](b \circ a)(x)[/tex] and simplify now.
[tex](b \circ a)(x)[/tex]
[tex]b(a(x))[/tex]
[tex]b(3x+1)[/tex]
[tex]\sqrt{(3x+1)-4}[/tex]
[tex]\sqrt{3x+1-4}[/tex]
[tex]\sqrt{3x-3}[/tex]
Now again we can only square root positive or zero numbers so we want [tex]3x-3 \ge 0[/tex].
Let's solve this to find the domain of [tex](b \circ a)(x)[/tex].
[tex]3x-3 \ge 0[/tex]
Add 3 on both sides:
[tex]3x \ge 3[/tex]
Divide both sides by 3:
[tex]x \ge 1[/tex]
So we want [tex]x[/tex] to be a number greater than or equal to 1.
The option that says this is [tex][1,\infty)[/tex]
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Give an example why option A fails:
A number in the given set is -2.
[tex]a(x)=3x+1[/tex]
[tex]b(x)=\sqrt{x-4}[/tex]
So [tex]a(-2)=3(-2)+1=-6+1=-5[/tex] and [tex]b(-5)=\sqrt{-5-4}=\sqrt{-9} \text{ which is not real}[/tex].
Give an example why option B fails:
A number in the given set is 0.
[tex]a(x)=3x+1[/tex]
[tex]b(x)=\sqrt{x-4}[/tex]
So [tex]a(0)=3(0)+1=0+1=1[/tex] and [tex]b(1)=\sqrt{1-4}=\sqrt{-3} \text{ which is not real}[/tex].
Give an example why option D fails:
While all the numbers in set D work, there are more numbers outside that range of numbers that also work.
A number not in the given set that works is 3.
[tex]a(x)=3x+1[/tex]
[tex]b(x)=\sqrt{x-4}[/tex]
So [tex]a(3)=3(3)+1=9+1=10[/tex] and [tex]b(1)=\sqrt{10-4}=\sqrt{6} \text{ which is real}[/tex].
