In Exercises 87–88, find a. (f ○ g)(x); b. the domain of (f ○ g)....

Question

In Exercises 87–88, find a. (f ○ g)(x); b. the domain of (f ○ g). f(x) = (x + 1)/(x - 2), g(x) = 1/x

Accepted Answer

Welcome back. I'm so glad you're here. We've got two functions and we are going to combine them to get a composite function. Our first function is F of X equals two X minus one times X plus five. And our second function is G of X equals three over X. Now let's look at how those are combined that G of X is second that F of X is first. We know that this is the same thing as F of G of X, meaning that second one, the G of X is the innermost function that's going to be applied to the outermost function. The F of X. All the G of X is get wedged in to the F of X wherever there's an X. So let's see what that looks like. Um We'll start copying down the F of X function. So that will start with the two and we ran into an X. That means we replace it with our G of X. So instead of two X it's two times three over X. Now we continue with our F of X function minus one and that's times O instead of X it's going to be three over X. R. G of X. And now resuming with our F of X function that's plus five, we can simplify that two times three over X. That's going to be six over X. And everything else just carries down. So six over X minus one, times three over X plus five. And now we can just spoil this. So our first six over X times three over X is going to yield 18 over x squared our outsides six over X times five is going to give us plus over X. Our insides negative one times three over X. That's going to give us minus three over X. And our last negative one times five is just going to be minus five. Will combine our like terms in the middle and we'll just carry down the 18 over X squared plus 30 over x minus three over X. S 27 over x minus five. This is our composite function F of G of X. Except we have to make sure there are no domain restrictions. So in order to look at the domain, we have to look at our original two functions F of X equals to x minus one times X plus five. There's no domain issues there. That's all real numbers. However, for G of X equals three over X. There is a domain issue there. Three over X. Well, X can't equal zero in the denominator. So we have to say that X cannot equal zero. Or in other words we could say that X can be anything from negative infinity up to zero but not including zero. United with zero. All the way up to positive infinity. We look at our answer choices with this domain and this function, this composite function and that is going to match with answer choice. D Well done. We'll catch you on the next one

In Exercises 87–88, find a. (f ○ g)(x); b. the domain of (f ○ g). f(x) = (x + 1)/(x - 2), g(x) = 1/x

Key Concepts

Function Composition

Domain of a Function

Rational Functions

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