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)= = (5x+1)/(x² - 9), g(x) = (4x -2)/(x² - 9)

Accepted Answer

Welcome back. I'm so glad you're here. We have two functions F of X and G of X. And we are to find F plus G and to write the domain. Alright for F plus G, what does that mean? We're going to take our F of X function and add it to our G of X function. R F of X function is nine X plus two divided by X squared minus 49. R G of X function is eight X plus nine divided by X squared minus 49. And as I mentioned before we just add them together. Well with these two functions we've got two fractions and to add two fractions you need to have a common denominator. Well they already do ye. So we can just add the too numerous to enumerators together. Nine X plus two plus eight X plus nine. And then combine our denominator. So X squared minus 49 remains our denominator. Let's combine like terms in the numerator. Nine X plus eight X will give us 17 X and two plus nine will give us plus 11. Still divided by X squared minus 49. We can factor that denominator but we can't factor the numerator. So there won't be any factors that cancel out between the numerator and the denominator. So we can leave it just like this. Alright, Part two. To write the domain for the domain. We have to look at our two original functions. So we look at a five X and G of X. And we want to see if there are any domain restrictions and yes, when we have a fraction, there's a domain restriction. What is it? Well, the denominator cannot be equal to zero. Otherwise it will be undefined. And because we have an X variable in the denominator, there are some values for X. That could make the denominator equal to zero. What are those? Well, we set the denominator equal to zero in order to find out what values for X would make this zero in the denominator and therefore undefined. And because they both have the same denominator, we only have to do this once. X squared minus 49 equals zero. Well, to factor that we notice that we've got a difference of squares we recall from previous lessons that for a difference of squares we take the square root of each of our two terms. So the squared of x squared, that's going to be X. And that goes in the first spot. In each set of parentheses, we look at the square root of the second number. The second number here is 49. The squared of 49 is seven. So that goes in the second spot of each set of parentheses. And then by definition one is plus one is minus. And then we are going to set each of these factors equal to zero to find out what values for X would give us zero in the denominator. Alright, for X plus seven equals zero. Let's subtract seven from both sides and that will leave us with X equals negative seven. And for X plus or x minus seven, let's add seven to both sides. And that will give us X equals seven. So if X equals negative seven or positive seven we would have a zero for the denominator and our functions would be undefined. So X cannot equal negative seven or positive seven. How do we express this in interval notation? Well, X can be any number under the sun starting from negative infinity. All the way up to negative seven and then it cannot equal negative seven. So we close that with the parentheses and we say that is in union with another parentheses starting just after negative seven. All the way up to what value? Up just before seven. It cannot be seven. So we put parentheses in union with and one more parentheses to say it can be any number just after seven. All the way up. And there are no more restrictions. So all the way up to infinity and we'll never quite obtain infinity. So we close infinity with the parentheses as well. 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 31–50, find ƒ+g and determine the domain for each function. f(x)= = (5x+1)/(x² - 9), g(x) = (4x -2)/(x² - 9)

Key Concepts

Function Addition

Domain of a Function

Rational Functions

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