Which function has a domain of [3, ∞)

Question

a. f(x) = √x-3

b. f(x) = √√x+3-3

c. f(x) = √x-3+8

d.f(x) = √x+3

Answer 1

When looking for the domain of an equation that has a square root, take a look at everything inside the square root first. You're looking for an equation where the smallest possible x value can be 3.

Automatically, we can see that D isn't what we're looking for. The smallest possible value inside that square root is -3. Moving on, let's take a look at B. Once again, we have x+3 inside the equation, but we have a secondary square root as well. Let's break this one down.

We want the square root of (x+3) to be equal to three, in order for the equation to equal 0. That means that (x+3) = 9 x=6. This doesn't work for our domain either.

Both A and C have the square root of x-3. Let's plug in 3 for equation A. f(x)= 0. We can't plug in a value of x that is smaller than 3, so our domain matches equation A.

Answer 2

a. sqr(x -3) has domain [3, infinity)

but c. also has the same domain

c. f(x) = sqr(x-3) + 8 has a domain x greater than or equal to 3, to positive infinity

but d could be correct if the answer there is (sqrx) +3 and not sqr(x+3)

where you seemingly have implicit parentheses makes a huge difference.

b looks like a misprint

answer(s) are a & c, or d

mostly likely "best" answers are a &c,

although hyper-technically the way the problem is written, d is the correct answer

For certain b is a wrong answer.