Question

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

Answer 1

Answer:

C.

Step-by-step explanation:

Answer 2

Answer:

$[1,\infty)$

Step-by-step explanation:

$b(x)=\sqrt{x-4}$

$a(x)=3x+1$

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.

$(b \circ a)(x)$

$b(a(x))$

$b(3x+1)$

$\sqrt{(3x+1)-4}$

$\sqrt{3x+1-4}$

$\sqrt{3x-3}$

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 $[1,\infty)$

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Give an example why option A fails:

A number in the given set is -2.

$a(x)=3x+1$

$b(x)=\sqrt{x-4}$

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

Give an example why option B fails:

A number in the given set is 0.

$a(x)=3x+1$

$b(x)=\sqrt{x-4}$

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

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.

$a(x)=3x+1$

$b(x)=\sqrt{x-4}$

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