In Exercises 31–50, find ƒ+g and determine the domain for each fu...

Question

In Exercises 31–50, find ƒ+g and determine the domain for each function. f(x)= = 8x/(x - 2), g(x) = 6/(x+3)

Accepted Answer

Welcome back. I'm so glad you're here. We are given two functions F of X and G of X. And we are to find F plus G and we are to write the domain. Well, I'm actually going to take a moment and work on the domain first are two functions F of X equals 18. X over X plus nine, and G of X equals nine, divided by x minus seven. We have domain restrictions here because these are fractions. And whenever we have a fraction we cannot have a zero in the denominator. That would make our fractions undefined. So we need to figure out what X values would give us a zero in the denominator. In order to do that, we set our denominators equal to zero. So X plus nine equals zero and X minus seven equals zero, solving for X will tell us what X values would give us a zero. So we subtract nine from both sides, which will give us an X cannot equal a negative nine. If X equals negative nine would have a zero in the denominator and it would be undefined looking at X minus seven equals zero. Likewise, we'll add seven to both sides and that will give us X equals seven, meaning X cannot equal seven. If x equals seven would have a zero in our denominator and it would be undefined. Okay, how do we express that in interval notation? Well, what we do is we start with negative infinity and we're saying X could be any real number. Starting all the way with negative infinity. All the way up to and we hit a bump at negative nine because it cannot be negative nine. So we put negative nine and a parentheses meaning but not negative nine. And that's in union with and we start again with negative nine because it's the numbers just after negative nine all the way up to the next bump in the road which is seven. And we close seven with the parentheses because we are saying it cannot be seven but it's all the way up to seven. And then parentheses starting just after seven and going all the way up to well we don't hit any more bumps so it's going all the way up to infinity. We close infinity with the parentheses because we'll never quite reach infinity. Okay, that is our domain. And now we can take a look at adding F plus G. Or F of X plus G of X. So when we have a five x 18 X over X plus nine plus G of X. Nine over X minus seven, we want to find a common denominator. So in order to do that, we're going to multiply each fraction by the other denominator over itself. So for 18 X over X plus nine the other denominator is x minus seven. And we put that over itself because that's just one for nine over X minus seven. The other denominator is X plus nine. We put that over itself because that's just one. And now we can go ahead and we can take X. Times X minus seven over X minus seven times X plus nine Plus nine times x plus nine over. Oops X plus x minus seven times X plus nine. And we have a common denominator. Now let's go ahead and write. So we have 18 X times X minus seven. We'll write this numerator on one line plus nine times X. Plus nine. And let's foil out our denominator. So first X. Times X. That'll be X squared R. Outside X. Times nine is plus nine X. Insides negative seven times X. Is minus seven X. And our last negative seven times nine is minus 63. Or if we combine our middle like terms here this will be X squared Plus two X -63. and I just want to pause before we keep going and simplify any further. Look at our answer choices. C. And D. Are kind of similar to this that's what our numerator looks like almost. But look the denominators none of them have that two X. In it in the middle. It's X. Squared in A. 16 X. So none of these work. We do not need to keep going down this road to figure out which of C. Or D. It would be. And so if we keep going what it look like A. Or B. No it would not because actually A. And B. That's just about what we started with. So let's look carefully at what we started with and see if it matches either A or B. Well, 18 X over X plus nine. That's the same for them. But then one is A plus and one is A minus. And yeah, we added the two so we can rule out B. And that just leaves us with answer choice A, which does match both solutions for F plus G and the domain. Well, well done, we will catch you on the next one.

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In Exercises 31–50, find ƒ+g and determine the domain for each function. f(x)= = 8x/(x - 2), g(x) = 6/(x+3)

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