Given the functions f(x)=1x2−2f(x)=\frac{1}{x^2-2} and g(x)=x+2g(x)=\sqrt{x+2}, find (f∘g)(x)(f\circ g)(x) and (g∘f)(x)(g\circ f)(x)

( f ∘ g ) ( x ) = 1 x (f\circ g)(x)=\frac{1}{x} ( f ∘ g ) ( x ) = x 1 ; ( g ∘ f ) ( x ) = 2 x 2 − 3 x 2 − 2 (g\circ f)(x)=\sqrt{\frac{2x^2-3}{x^2-2}} ( g ∘ f ) ( x ) = x 2 − 2 2 x 2 − 3 <path d="M983 90 l0 -0 c4,-6.7,10,-10,18,-10 H400000v40 H1013.1s-83.4,268,-264.1,840c-180.7,572,-277,876.3,-289,913c-4.7,4.7,-12.7,7,-24,7 s-12,0,-12,0c-1.3,-3.3,-3.7,-11.7,-7,-25c-35.3,-125.3,-106.7,-373.3,-214,-744 c-10,12,-21,25,-33,39s-32,39,-32,39c-6,-5.3,-15,-14,-27,-26s25,-30,25,-30 c26.7,-32.7,52,-63,76,-91s52,-60,52,-60s208,722,208,722 c56,-175.3,126.3,-397.3,211,-666c84.7,-268.7,153.8,-488.2,207.5,-658.5 c53.7,-170.3,84.5,-266.8,92.5,-289.5z M1001 80h400000v40h-400000z">

Given the functions f ( x ) = 1 x 2 − 2 f(x)=\frac{1}{x^2-2} and g ( x ) = x + 2 g(x)=\sqrt{x+2} , find ( f ∘ g ) ( x ) (f\circ g)(x) and ( g ∘ f ) ( x ) (g\circ f)(x)

Hey, everyone, and welcome back. So let's see if we can solve this problem right here. In this problem, we are given 2 functions. We're given f of x is equal to x squared minus 2, or excuse me, one over x squared minus 2, and g of x is equal to the square root of x plus 2. And we're asked to find f of g of x, and then we're asked to find g of f of x. So to solve this problem, we're first going to find f of g of x where g is the inside function. This is part a. And to do this, we're going to take our g and we're going to plug it in to our function f of x. So what we're gonna do is replace this x with what we have for g. So that means our function f of x is going to become or f of g of x, I should say, would become one over the square root of x plus 2 squared. Notice how this x just got replaced with the function for g, and then this whole thing will be minus 2. Now Now at this point, I can simplify this function, because notice that the square root and the square will cancel, giving us that this whole thing is equal to 1 over x+2-2, and at this point, I can cancel the twos here, giving me that this whole function f of g of x is equal to 1 over x. So when we compose this function f of g of x, we end up getting one over x. So that is how you can compose this by putting g inside of f. But now let's see if we can do part b, where we're asked to put f inside of g. So we'll have g of f of x. And to do this part, we're just going to take our function for f of x and put it inside of g of x. So this is what we're going to end up with. So what we can do for this is take this x and replace it with this whole function, meaning we're going to have the square root of 1 over x squared minus 2, that's what we replace this with, plus 2. Now it's kind of difficult to simplify from here, but one of the things we could do is actually try to get this all as one fraction by making the denominator similar, because 2 is really the same thing as 2 over 1. So let's see if we can do this. Well, to start, we can keep 1 over x squared minus 2, but what I'm going to do here is I'm going to take this fraction, and I'm going to multiply the top and bottom by this denominator. So we'll have 2 times x squared minus 2 divided by x squared minus 2. Notice now we have similar denominators, so we can combine these two fractions. So if I add the tops together, what I can do is I'll get 1 +2 times x squared minus 2, and then this whole thing is divided by x squared minus 2. Now from here, what I can do is I can also take this 2 and distribute it into each of these terms inside. So this is going to give me the square root of 1 plus, and then 2 times x squared is 2x squared, and then 2 times negative 2 is going to give us negative 4, then this whole thing is divided by x squared minus 2. Now the last step that I can take here is I can is I can combine the 1 and the negative 4, which will give me negative 3, because one plus negative 4 is negative 3. So then this whole thing will be the square root of 2x squared minus 3, because we have a negative 3 all divided by x squared minus 2. So this right here would be the most simplified answer that we can get for g of f of x. When we compose the functions in this way, we get this as our most simplified answer. So that is how you can compose other functions. These were some more examples. Hope you found this helpful. Thanks for watching.

Given the functions f ( x ) = 1 x 2 − 2 f(x)=\frac{1}{x^2-2} and g ( x ) = x + 2 g(x)=\sqrt{x+2} , find ( f ∘ g ) ( x ) (f\circ g)(x) and ( g ∘ f ) ( x ) (g\circ f)(x)

( f ∘ g ) ( x ) = 1 x (f\circ g)(x)=\frac{1}{x} ( f ∘ g ) ( x ) = x 1 ; ( g ∘ f ) ( x ) = 2 x 2 − 3 x 2 − 2 (g\circ f)(x)=\sqrt{\frac{2x^2-3}{x^2-2}} ( g ∘ f ) ( x ) = x 2 − 2 2 x 2 − 3 <path d="M983 90 l0 -0 c4,-6.7,10,-10,18,-10 H400000v40 H1013.1s-83.4,268,-264.1,840c-180.7,572,-277,876.3,-289,913c-4.7,4.7,-12.7,7,-24,7 s-12,0,-12,0c-1.3,-3.3,-3.7,-11.7,-7,-25c-35.3,-125.3,-106.7,-373.3,-214,-744 c-10,12,-21,25,-33,39s-32,39,-32,39c-6,-5.3,-15,-14,-27,-26s25,-30,25,-30 c26.7,-32.7,52,-63,76,-91s52,-60,52,-60s208,722,208,722 c56,-175.3,126.3,-397.3,211,-666c84.7,-268.7,153.8,-488.2,207.5,-658.5 c53.7,-170.3,84.5,-266.8,92.5,-289.5z M1001 80h400000v40h-400000z">

( f ∘ g ) ( x ) = 1 x (f\circ g)(x)=\frac{1}{x} ( f ∘ g ) ( x ) = x 1 ; ( g ∘ f ) ( x ) = 3 x 2 − 2 (g\circ f)(x)=\sqrt{\frac{3}{x^2-2}} ( g ∘ f ) ( x ) = x 2 − 2 3 <path d="M983 90 l0 -0 c4,-6.7,10,-10,18,-10 H400000v40 H1013.1s-83.4,268,-264.1,840c-180.7,572,-277,876.3,-289,913c-4.7,4.7,-12.7,7,-24,7 s-12,0,-12,0c-1.3,-3.3,-3.7,-11.7,-7,-25c-35.3,-125.3,-106.7,-373.3,-214,-744 c-10,12,-21,25,-33,39s-32,39,-32,39c-6,-5.3,-15,-14,-27,-26s25,-30,25,-30 c26.7,-32.7,52,-63,76,-91s52,-60,52,-60s208,722,208,722 c56,-175.3,126.3,-397.3,211,-666c84.7,-268.7,153.8,-488.2,207.5,-658.5 c53.7,-170.3,84.5,-266.8,92.5,-289.5z M1001 80h400000v40h-400000z">

( f ∘ g ) ( x ) = x (f∘g)(x)=x ( f ∘ g ) ( x ) = x ; ( g ∘ f ) ( x ) = x 2 − 2 (g\circ f)(x)=\sqrt{x^2-2} ( g ∘ f ) ( x ) = x 2 − 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

( f ∘ g ) ( x ) = x (f∘g)(x)=x ( f ∘ g ) ( x ) = x ; ( g ∘ f ) ( x ) = 1 x + 2 − 2 (g\circ f)(x)=\frac{1}{\sqrt{x+2}-2} ( g ∘ f ) ( x ) = x + 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> − 2 1

10. Relations and Functions

Composition of Functions

Beginning & Intermediate Algebra

10. Relations and Functions

Composition of Functions

Struggling with Beginning & Intermediate Algebra?

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Multiple Choice

Given the functions $f (x) = \frac{1}{x^{2} - 2}$ and $g (x) = \sqrt{x + 2}$ , find $(f \circ g) (x)$ and $(g \circ f) (x)$

$(f$\circ$ g)(x)=$\frac{1}{x}$$ ; $(g$\circ$ f)(x)=$\sqrt{\frac{2x^2-3}{x^2-2}$}$

$(f$\circ$ g)(x)=$\frac{1}{x}$$ ; $(g$\circ$ f)(x)=$\sqrt{\frac{3}{x^2-2}$}$

$(f∘g)(x)=x$ ; $(g$\circ$ f)(x)=$\sqrt{x^2-2}$$

$(f∘g)(x)=x$ ; $(g$\circ$ f)(x)=$\frac{1}{\sqrt{x+2}$-2}$

Verified step by step guidance

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) = $\frac{1}{x^2 - 2}$ and g(x) = $\sqrt{x + 2}$, substitute g(x) into f: replace every x in f(x) with g(x). This gives f(g(x)) = $\frac{1}{(\sqrt{x + 2}$)^2 - 2}.

Simplify the expression inside the denominator: ($\sqrt{x + 2}$)^2 simplifies to x + 2, so the denominator becomes (x + 2) - 2.

Simplify the denominator further: (x + 2) - 2 simplifies to x. So, (f $\circ$ g)(x) = $\frac{1}{x}$.

Next, find (g $\circ$ f)(x) = g(f(x)) by substituting f(x) into g. Since g(x) = $\sqrt{x + 2}$, replace x with f(x) to get g(f(x)) = $\sqrt{\frac{1}{x^2 - 2}$ + 2}. Then combine the terms inside the square root by expressing 2 as $\frac{2(x^2 - 2)}{x^2 - 2}$ to have a common denominator, and simplify the numerator accordingly.

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Struggling with Beginning & Intermediate Algebra?

Given the functions f(x)=1x2−2f(x)=\(\frac{1}{x^2-2}\) and g(x)=x+2g(x)=\(\sqrt{x+2}\), find (f∘g)(x)(f\(\circ\) g)(x) and (g∘f)(x)(g\(\circ\) f)(x)

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Given the functions $f (x) = \frac{1}{x^{2} - 2}$ and $g (x) = \sqrt{x + 2}$ , find $(f \circ g) (x)$ and $(g \circ f) (x)$