Question

If a(x) = 3x + 1 and b (x) = StartRoot x minus 4 EndRoot, what is the domain of (b circle a) (x)?

(negative infinity, infinity)
Left-bracket 0, infinity)
Left-bracket 1, infinity)
Left-bracket 4, infinity)

Answer 1

C

Step-by-step explanation:

Hope you have a great day!

Answer 2

`[1,\infty)`

Step-by-step explanation:

`b(x)=√(x-4)`

`a(x)=3x+1`

Since we want to know the domain of
`(b \circ a)(x)`, let's first consider the domain of the inside function, that is, that of
`a(x)=3x+1`. Every polynomial function has domain all real numbers.

So we can plug anything for function
`a` 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
`(b \circ a)(x)` and simplify now.

`(b \circ a)(x)`

`b(a(x))`

`b(3x+1)`

`√((3x+1)-4)`

`√(3x+1-4)`

`√(3x-3)`

Now again we can only square root positive or zero numbers so we want
`3x-3 \ge 0`.

Let's solve this to find the domain of
`(b \circ a)(x)`.

`3x-3 \ge 0`

Add 3 on both sides:

`3x \ge 3`

Divide both sides by 3:

`x \ge 1`

So we want
`x` to be a number greater than or equal to 1.

The option that says this is
`[1,\infty)`

-------------------------------

Give an example why option A fails:

A number in the given set is -2.

`a(x)=3x+1`

`b(x)=√(x-4)`

So
`a(-2)=3(-2)+1=-6+1=-5` and
`b(-5)=√(-5-4)=√(-9) \text{ which is not real}`.

Give an example why option B fails:

A number in the given set is 0.

`a(x)=3x+1`

`b(x)=√(x-4)`

So
`a(0)=3(0)+1=0+1=1` and
`b(1)=√(1-4)=√(-3) \text{ which is not real}`.

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.

`a(x)=3x+1`

`b(x)=√(x-4)`

So
`a(3)=3(3)+1=9+1=10` and
`b(1)=√(10-4)=√(6) \text{ which is real}`.