In Problems 23-38, for the given functions f and g, find: (a) f ∘g (b) g ∘f (c) f ∘f (d) g ∘g St

In Problems 23-38, for the given functions f and g, find:...

Key Concepts

Rational Functions

Rational functions are functions expressed as the ratio of two polynomials. They have domains defined by all real numbers except those that make the denominator zero, leading to undefined expressions. When working with rational functions in compositions, it is important to identify and exclude values that result in a zero denominator in either the inner or outer function.

Function Composition

Function composition involves applying one function to the results of another, denoted as (f ? g)(x) = f(g(x)). This process requires substituting the entire expression of the inner function g(x) into every occurrence of x in the outer function f(x), and then simplifying if needed. It is a fundamental operation that allows the creation of new functions from existing ones.

Domain Restrictions in Composite Functions

When dealing with composite functions, it is crucial to consider the domain of each function separately. The domain of the composite function f ? g consists of all x-values that are in the domain of g and that yield values through g(x) which are in the domain of f. This means any restrictions from both functions, such as values that would cause division by zero or other undefined expressions, must be taken into account.

Transcript

00:01 So for this question, there are two different functions, and there are four different questions, actually, within this function.

00:08 The first, they're asking us to do the composition of f and g.

00:13 So basically, it's like f of g of x, meaning g of x goes in for all of the x's within the f function.

00:25 So our f function was x minus 5 x plus 1 our g function was x plus 2 over x minus 3.

00:53 So i'm going to take my f function, i'm going to plug in the g for each of the x's.

01:01 So this whole thing is essentially going in for x here and x here.

01:07 So i like to write it like this.

01:12 I've got x plus 2 over x minus 3 minus 5 all over x plus 2 over x minus 3 plus 1.

01:27 I simplify the top and the bottom first before i do the division.

01:31 I do that by multiplying the top and the bottom of this by x minus 3.

01:38 Same thing down here.

01:40 I multiply the top and the bottom by x minus 3.

01:44 3.

01:46 That's going to give me x plus 2 minus 5x plus 15.

01:54 I'm distributing this here all over x minus 3.

02:01 That is all over x plus 2 plus x minus 3 all over again x minus 3.

02:12 Now we know because this is a division problem.

02:17 If i write it out, i'll simplify for this one.

02:20 The other ones i won't do.

02:21 I'll do a shorthand version.

02:23 I get negative 4x plus 17 over x minus 3.

02:31 And then on the bottom here, i've got 2x minus 1 over x minus 3.

02:41 So when we physically write out the division problem, it's negative 4x plus 17 over x minus 3 divided by 2x minus 1 over x minus 3.

02:54 We know that division is really multiplying by the reciprocal.

02:58 So negative 4x plus 17 over x minus 3 multiplied by x minus 3 over 2x minus 1.

03:09 So these we're going to flip, right? now when i go to multiply these, i can cross can't.

03:15 Cancel early my x minus 3s.

03:18 So i just have negative 4x plus 17 over the 2x minus 1.

03:25 And that is going to be our answer for a in this question...

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