For each pair of functions f and g below, find f(g(x)) and...

For each pair of functions \(f\) and \(g\) below, find \(f(g(x))\) and \(g(f(x))\). Then, determine whether \(f\) and \(g\) are inverses of each other. Simplify your answers as much as possible. (Assume that your expressions are defined for all \(x\) in the domain of the composition. You do not have to indicate the domain.) (a) \(f(x) = 4x\) \(g(x) = \frac{x}{4}\) \(f(g(x)) = \) \(g(f(x)) = \) \(\circ f\) and \(g\) are inverses of each other \(\circ f\) and \(g\) are not inverses of each other (b) \(f(x) = 2x + 7\) \(g(x) = \frac{x - 7}{2}\) \(f(g(x)) = \) \(g(f(x)) = \) \(\circ f\) and \(g\) are inverses of each other \(\circ f\) and \(g\) are not inverses of each other

Transcript

00:01 So in the first problem, we're given that f of x is equal to 3 over x, and that g of x is also equal to 3 over x.

00:06 We want to find f of g of x and g of f of x.

00:10 So let's start by finding f of g of x.

00:12 So remember, we always start with our inside function, which is g of x.

00:16 So that means we would have f of 3 over x.

00:20 Well, to find f of 3 over x, we're going to substitute 3 over x in place of x in our f function, which leaves us with 3 divided by 3 over x.

00:30 So one thing we can do in order to get rid of that denominator within the denominator, i can multiply the numerator and denominator by our denominator.

00:38 So i'm going to multiply by x over x.

00:40 So for our numerators, 3 times x would be 3x.

00:43 But for 3 over x times x, those x's will cancel out, which just leaves us with 3.

00:49 And 3x divided by 3 is equal to x.

00:52 So f of g of x is equal to x.

00:55 Now for the second part of this, we want to find g of f of x.

00:59 So again, we're going to start with our inside function, which is f of x, which means we're going to have g of 3 over x.

01:06 Well, now we need to substitute 3 over x in place of x in our g function.

01:10 Well, that means we're going to have, again, 3 divided by 3 over x.

01:14 So we're going to solve in the same way.

01:16 We can multiply the numerator and denominator by x.

01:19 3 times x is 3x.

01:21 And 3 over x times x is 3.

01:23 And 3x divided by 3 is equal to x, which means that g of f of x is also x.

01:29 So now that we've found that both f of g of x and g of f of x both equal to x, that's how we know they are inverses of each other.

01:38 So yes, in this case, they are inverses...

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