In Exercises 1–30, find the domain of each function. f(x) = 1/√(x...

Question

In Exercises 1–30, find the domain of each function. f(x) = 1/√(x - 3)

Accepted Answer

Welcome back. I'm so glad you're here. We have the function F of X equals six divided by the square root of four. X minus 12. We are asked to find the domain. Well, we've got two domain restrictions here. One is because we have a fraction and when we have a fraction, the denominator cannot equal zero. The other restriction is that we have this radical here. And what's inside the radical must be greater than or equal to zero. So let's work on both of those. Okay, the denominator cannot equal zero. Why not? Well, if we have a zero in the denominator, it's going to be undefined And we don't want it to be undefined. So to find out what values of X would make the denominator equal to zero. And therefore undefined. We're going to set that entire denominator. The square root of four, X -12 equal to zero. And then when we solve for X, that will tell us what values would give us a zero in the denominator. Let's go ahead and do that work. We are going to square both sides here and the square root of four, X minus 12 squared is now four x minus and zero square that's still zero. We'll add 12 to both sides. We have four. X equals 12. We'll divide both sides by four and we get X equals three. What does that mean? Well, if X equals three, then our denominator is equal to zero and we just can't have that. That would be undefined. Therefore X cannot equal three. There's one of our domain restrictions. The other one I mentioned before, is that what's inside the radical? Or that four X minus 12 must be greater than or equal to zero? Y. Because if that whole thing is less than zero, if it's a negative number then we're taking the square root of a negative which is imaginary. And we don't want to deal with imaginary numbers for our domain. So we set that for X minus 12 greater than or equal to zero and we are going to solve for X. So we've got adding 12 double sides, we have four. X is greater than or equal to 12 will divide both sides by four and we'll get that X must be greater than or equal to three. But wait a second. Looking at our other one. X cannot equal three because that would give us a denominator of zero. So it's not X is greater than equal to three in this case. Because of our other domain restriction, it's just X is greater than three. Let's draw this on a number line. If we're going from negative infinity up to positive infinity of zero. Somewhere in the middle X must be greater than three. Well, here's three. It's an open circle because it can't be three and it's going all the way up to positive infinity because X has to be greater than three. How do we write that in interval notation? We're going to use a parentheses for the first value that we can almost have. So it's parentheses three, meaning not including three, but everything just after three. All the way up to positive infinity. Closed with a parentheses because we can't ever quite obtain infinity. We look at our answer choices and this matches with answer choice B. Well done, we'll catch you on the next one.

In Exercises 1–30, find the domain of each function. f(x) = 1/√(x - 3)

Key Concepts

Domain of a Function

Square Root Function

Inequalities

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