Question

State the domain of the following function, in interval notation.
f(x)=(3x+7)^1/2

Answer 1

We know \[\sqrt{g(x)} \not \in \mathbb{R} \] when \[g(x)<0\] so we find when \[g(x) \geq 0 \]
Which is: \[3x+7 \geq 0\] Solving this, we receive: \[x \geq -\frac{7}{3}\] Which implies \[x \in \left(-\frac{7}{3}, \infty\right)\] And that is our domain.

Answer 2

Wait, wait, oops, \[x \in \left[-\frac{7}{3}, \infty\right)\]

Answer 3

Wow, so it's (-inf, -7/3) U (-7/3, inf) ?

Answer 4

It's the latter, but it's finitely bounded below, because \[x^\frac{1}{2}=\sqrt{x}\] so it's only a real number when it is [-7/3, inf)

Answer 5

So you had to get rid of the exponent by taking the square root?

Answer 6

Careful, you have to remember that \[(n^\frac{1}{2})^2=n\] right? But also remember that's the definition of a square root. "The square root of n is a number such that when squared, is n."
So, we can say that \[n^\frac{1}{2}\] is actually *equivalent* to \[\sqrt{n}\]They are the same thing.

Answer 7

Correct. I was so confused! It always gets me, too, because sometimes, it's hard to tell when something is "greater/less than or equal to" or "greater/less than." I think I got it now, though!

Answer 8

Haha, awesome, PM me if you have any more questions.

Answer 9

Thank you so much!!!