Given the functions f(x)=x+4f\left(x\right)=\sqrt{x+4} and g(x)=(x−2)2−4g\left(x\right)=\left(x-2\right)^2-4, find (f∘g)(x)\left(f\circ g\right)\left(x\right) and (g∘f)(x)\left(g\circ f\right)\left(x\right).

(f∘g)(x)=x−2\left(f\circ g\right)\left(x\right)=x-2(f∘g)(x)=x−2 ; (g∘f)(x)=(x+4)−4x+4\left(g\circ f\right)\left(x\right)=\left(x+4\right)-4\sqrt{x+4}(g∘f)(x)=(x+4)−4x+4

Given the functions f ( x ) = x + 4 f\left(x\right)=\sqrt{x+4} and g ( x ) = ( x − 2 ) 2 − 4 g\left(x\right)=\left(x-2\right)^2-4 , find ( f ∘ g ) ( x ) \left(f\circ g\right)\left(x\right) and ( g ∘ f ) ( x ) \left(g\circ f\right)\left(x\right) .

Alright. So let's give this problem a try. In this problem, we have 2 functions. We have f of x is equal to the square root of x plus 4, and g of x is equal to x minus 2 squared minus 4. For part a of this problem, we're asked to find f of g of x, and for part b, we're asked to find g of f of x. So let's see how we can solve this. First off, I'll do part a, which is where we're finding f of g of x. And recall that the second letter that you see here is the inside function. Now to do this, we're going to take our function for g of x and put it inside of f of x. So what's gonna end up happening is we'll get f of x, which is the square root, and we have the x plus 4, but we're going to replace this x with everything we have for g of x. So we're going to have x minus 2 squared minus 4, and then this whole thing will be plus 4 because we have this plus 4 on the outside here. Now what I can do from here is simplify this function. So to simplify this, I can see that the minus 4 and the plus 4 will cancel, giving us the square root of x minus 2 squared. And from here, I can cancel the square and the square root, giving me x minus 2. So when we compose f and g, the final function that we get for f of g of x looks like this. But now let's see how we can solve part b of this problem. In part b, we're asked to find g of f of x. Now to do this, we're gonna go backwards and take f of x and plug it inside of g. So to do this, I'm gonna take this entire thing, the square root of x plus 4, and we'll plug it into the x in g of x up here. And let me actually clear some of these markings just because this is getting kind of confusing up here. So to do this, we're going to have our function, g of x, which is x minus 2 squared minus 4. We're gonna take this x and replace it with the function we have here. So x is going to be replaced with the square root of x+4, and then for the rest of this function, we have minus 2 because that's part of the g of x function. This whole thing is squared, and then we have this whole thing minus 4. Now to simplify this, I'm gonna go ahead and foil out this situation right here, because x the square root of x the square root of x plus 4 minus 2 squared, this whole thing can be written as square root of x+4 minus 2 times the square root of x+4minus2, and then this whole thing, of course, will be subtracted from 4. Now to do this, we can go ahead and just use the foil method, so I'll multiply these 2 together. The square root of x plus 4 times the square root of x plus 4 is just going to give us x+4. This is the first term that we get. Now the next thing I can do is take the square root of x+4multiplyitheregivingusminus2 times the square root of x+4, then I can multiply the negative 2 times this giving us also minus 2 times the square root of x plus 4. And then we have negative 2 times negative 2, which will give us positive 4, and then this whole thing will be minus 4. Now if I look down here, I can see that these are going to combine into 1. So we're going to have x+4, yeah, x+4, and then we'll have minus and then a minus 2 times the quantity, minus another 2 times the quantity, that'll give us minus 4 times the square root of x+4. So I'm just combining these 2 zeros, subtracting them. And then we'll have plus 4 minus 4, and the 4 minus fours are are actually going to cancel, so these will cancel each other out. So our simplified function is going to be x+minuteus4 times the square root of x+4, and this right here would be the fully simplified answer for g of f of x. So when we take f and we plug it into g, this is the answer that we get. So this is how you can compose functions. These are some more examples. Hopefully, you found this helpful, and thanks for watching.

Given the functions f(x)=x+4f\left(x\right)=\sqrt{x+4} and g(x)=(x−2)2−4g\left(x\right)=\left(x-2\$$right)^2-4$$, find (f∘g)(x)\left(f\circ g\right)\left(x\right) and (g∘f)(x)\left(g\circ f\right)\left(x\right).

(f∘g)(x)=x−2\left(f\circ g\right)\left(x\right)=x-2(f∘g)(x)=x−2 ; (g∘f)(x)=(x+4)−4x+4\left(g\circ f\right)\left(x\right)=\left(x+4\right)-4\sqrt{x+4}(g∘f)(x)=(x+4)−4x+4

(f∘g)(x)=x−2\left(f\circ g\right)\left(x\right)=\sqrt{x-2}(f∘g)(x)=x−2 ; (g∘f)(x)=(x+4)−4x+4\left(g\circ f\right)\left(x\right)=\left(x+4\right)-4\sqrt{x+4}(g∘f)(x)=(x+4)−4x+4

(f∘g)(x)=x−2\left(f\circ g\right)\left(x\right)=x-2(f∘g)(x)=x−2 ; (g∘f)(x)=x(x+4)\left(g\circ f\right)\left(x\right)=x\left(x+4\right)(g∘f)(x)=x(x+4)

(f∘g)(x)=x−2\left(f\circ g\right)\left(x\right)=x-2(f∘g)(x)=x−2 ; (g∘f)(x)=4x−4\left(g\circ f\right)\left(x\right)=4\sqrt{x-4}(g∘f)(x)=4x−4

4. Graphs and Functions

Composition of Functions

Intermediate Algebra

4. Graphs and Functions

Composition of Functions

Intermediate Algebra 때문에 힘들어하고 계신가요?

저희를 신뢰하는 수천 명의 학생들이 시험 합류하세요!

객관식

Given the functions $f (x) = \sqrt{x + 4}$ and $g (x) = {(x - 2)}^{2} - 4$ , find $(f \circ g) (x)$ and $(g \circ f) (x)$ .

$\left\)(f$\circ$ g$\right$)$\left$(x$\right$)=$\sqrt{x-2}$ ; $\left$(g$\circ$ f$\right$)$\left$(x$\right$)=$\left$(x+4$\right$)-4\(\sqrt{x+4}$

$$\left$(f$\circ$ g$\right$)$\left$(x$\right$)=x-2$ ; $$\left$(g$\circ$ f$\right$)$\left$(x$\right$)=x$\left$(x+4$\right$)$

$$\left$(f$\circ$ g$\right$)$\left$(x$\right$)=x-2$ ; $\left\)(g$\circ$ f$\right$)$\left$(x$\right$)=4\(\sqrt{x-4}$

$$\left$(f$\circ$ g$\right$)$\left$(x$\right$)=x-2$ ; $\left\)(g$\circ$ f$\right$)$\left$(x$\right$)=$\left$(x+4$\right$)-4\(\sqrt{x+4}$

검증된 단계별 안내

Recall that the composition of functions $ (f \circ g)(x) $ means you substitute $ g(x) $ into the function $ f $. So, start by writing $ (f \circ g)(x) = f(g(x)) $.

Given $ f(x) = \sqrt{x + 4} $ and $ g(x) = (x - 2)^2 - 4 $, substitute $ g(x) $ into $ f $: $ f(g(x)) = \sqrt{g(x) + 4} = \sqrt{(x - 2)^2 - 4 + 4} $.

Simplify inside the square root: $ (x - 2)^2 - 4 + 4 = (x - 2)^2 $. So, $ (f \circ g)(x) = \sqrt{(x - 2)^2} $.

Recall that $ \sqrt{a^2} = |a| $, so $ \sqrt{(x - 2)^2} = |x - 2| $. Depending on the domain, this can simplify further to $ x - 2 $ if $ x \geq 2 $.

Next, for $ (g \circ f)(x) = g(f(x)) $, substitute $ f(x) $ into $ g $: $ g(f(x)) = (f(x) - 2)^2 - 4 = (\sqrt{x + 4} - 2)^2 - 4 $. Expand the square and simplify to express the composition in terms of $ x $.

4:56m

다음 시청

마스터 Function Compositions가 Patrick Ford 짧은 영상으로 설명해 줍니다

Composition of Functions 연습 세트

수석 튜터가 만든 문제 세트

전문가 영상 설명

Intermediate Algebra 때문에 힘들어하고 계신가요?

Given the functions f(x)=x+4f\(\left\)(x\(\right\))=\(\sqrt{x+4}\) and g(x)=(x−2)2−4g\(\left\)(x\(\right\))=\(\left\)(x-2\(\right\))^2-4, find (f∘g)(x)\(\left\)(f\(\circ\) g\(\right\))\(\left\)(x\(\right\)) and (g∘f)(x)\(\left\)(g\(\circ\) f\(\right\))\(\left\)(x\(\right\)).

다음 시청

Composition of Functions 연습 세트

Given the functions $f (x) = \sqrt{x + 4}$ and $g (x) = {(x - 2)}^{2} - 4$ , find $(f \circ g) (x)$ and $(g \circ f) (x)$ .