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

Question

In Exercises 31–50, find f/g and determine the domain for each function. f(x)= = (5x+1)/(x² - 9), g(x) = (4x -2)/(x² - 9)

Accepted Answer

Hello. Everyone in this video. We're going to be looking at this practice problem where we want to find F divided by G. And write the domain of F divided by G. And were given that F of X is equal to nine X plus two divided by X squared minus 49. And G of X is equal to eight X plus nine divided by x squared minus 49. So the first step that we can do is to find F divided by G and F divided by G is the same as the F function F F X divided by the G function G of X. So we are given values for F of X and G of X. So I can go ahead and plug those in. So F of X is equal to nine X plus two divided by X squared minus 49. And that function F of X is divided by G of X which is equal to eight x plus nine divided by X squared minus 49. So because I have fractions on both the numerator and the denominator, I'm going to rewrite this in a linear format so I can convert it to a multiplication. So if I convert it to a linear format, I'll have my first fraction nine X plus two divided by X squared minus 49. And that's being divided by my second fraction which is G of X or eight X plus nine divided by X squared minus 49. So in order to convert this to a multiplication to make it easier to simplify, I need to switch the numerator and denominator of my second fraction. So x squared - will be my new numerator. And my denominator will now be eight x plus nine. And you can see that now I have X squared minus 49 in the denominator of my first fraction and X squared minus 49 in the numerator of my second fraction. So these actually cancel out and become one. So when I do the multiplication I'm left with nine x plus two Divided by eight x plus nine times one. Which means that F divided by G. Is actually just equal to nine X plus two Divided by eight x plus nine. So now that I figured out what F divided by G is I need to find the domain of F divided by G. So in this case because we're working with a fraction, I need to make sure that my denominator is never equal to zero because if my denominator is equal to zero, the answer will be undefined. And I want to make sure that the X. Values that are included in the domain give me valid answers. So in this case the denominator is eight X plus nine. So I want to make sure that the denominator never equal zero. So I can go ahead and set an equation where I have eight X plus nine cannot equal zero. And if I sold for that I subtract nine from both sides That leaves me with eight X is equal to -9. And in order to isolate x all divide by eight on both sides. So that leaves me with X cannot equal negative 9/8. So now I have this value that I know X cannot equal. So I can build an interval for that. So X can go from negative infinity 2 -9/8 And I'll do a parentheses because I know that X cannot equal negative 9/8 exactly. But it can go past negative 9/8 and into infinity. So this becomes the range of X values that give me a defined answer for my equation F divided by G. And you might think that this is the domain but you have to remember that we did a composite of old functions. So because we did a composite, we have to make sure that the X values that are undefined in the old functions are still undefined in this new function. Basically, we need to make sure that we include undefined values in our new function. So if we look at the old functions, they're both fractions. So once again, we just want to make sure that the denominator does not equal zero. And in this case both F of X and G of X have the same denominator of X squared minus 49. So we just want to make sure that X squared -49 does not equal zero. So we can solve for this if I add 49 to both sides, I have X squared is cannot equal 49. And in order to get X by itself I need to do the square root of both sides. And that leaves me with X cannot equal plus or minus seven. Notice how he did plus or minus because the square root of 49 could be positive seven or negative seven. And now I need to write this equation into an interval. So for this one I can go from negative infinity to negative seven and then from -7 to positive seven and then from positive seven to positive infinity. So you can see that these interval ranges are different. So in order to figure out our domain for our composite function f divided by G. We need to combine these intervals. So I'm gonna go ahead and do a number line to make it easier to visualize these ranges. So I'm gonna say that this is zero, this is negative 9/8 positive 9/ positive seven, -7. And these can be just some number in between like -5 and positive five. Notice my Number line is not drawn to scale but let's go ahead and look at this first interval where we have negative infinity to negative 9/8. So that would go parentheses here and to the left And the -9/8 to positive infinity. So another parentheses the other way and an error going to the right. So that's the first interval drawn. Let's look at the second interval negative infinity to negative seven with negative seven to positive seven and seven to positive infinity. So the first one is Down here from -7 to negative infinity And then from - two positive seven. And then again from positive seven to positive infinity. So now we need to write the new interval range. So if I start from here I can go from negative infinity to negative seven and then we need to cut it because X cannot equal negative seven but I'll do a union from negative seven two negative 9/8. Again you can see a cut here because X cannot equal negative 9/8. And then another union From -9/ to positive seven. Where I see my next cut And then my final union from 7 to positive infinity. So now this becomes the interval range of X values. That give me a valid answer for the composite function F divided by G. Or just the domain. So you can see that this domain excludes any X values that results in the denominators being equal to zero. So this becomes my final answer for the domain of the function F divided by G. And F divided by G is nine X plus two divided by eight X plus nine. And you can see that both F divided by G. And the domain align with answer choice A. So hopefully this video helped and see you next time.

In Exercises 31–50, find f/g and determine the domain for each function. f(x)= = (5x+1)/(x² - 9), g(x) = (4x -2)/(x² - 9)

Key Concepts

Rational Functions

Domain of a Function

Quotient of Functions

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