# [f o g] [x] and [g o f] [x]? (Algebra 2)?

f(x)=x+5

g(x)=x^2+6

f(x)=2x

g(x)=3x-4

f(x)=x-4

g(x)=3x^2

I don't understand these. My book does a really bad job of explaining them. I remember that I have to use substitution however whenever I try to solve them that way it doesn't look right.

Thanks in advance!

*In class we do not use SIN, COS, and TAN.*

### 6 Answers

- 1 decade agoFavorite Answer
PROBLEM ONE.

f(x) = x + 5

g(x) = x^2 + 6

[f o g] (x) = f(g(x)) = f(x^2 + 6) = x^2 + 6 + 5 = x^2 + 11

[g o f] (x) = g(f(x)) = g(x + 5) = (x + 5)^2 + 6 = x^2 + 10x + 25 + 6 = x^2 + 10x + 31

PROBLEM TWO.

f(x) = 2x

g(x) = 3x - 4

[f o g] (x) = f(g(x)) = f(3x - 4) = 2(3x - 4) = 6x - 8

[g o f] (x) = g(f(x)) = g(2x) = 3(2x) - 4 = 6x - 4

PROBLEM THREE.

f(x) = x - 4

g(x) = 3x^2

[f o g] (x) = f(g(x)) = f(3x^2) = 3x^2 - 4

[g o f] (x) = g(f(x)) = g(x - 4) = 3(x-4)^2 = 3(x^2 - 8x + 16) = 3x^2 - 24x + 48

Note: You are not really solving anything here, you are just plugging things in in order to see what comes out. Also note that generally [f o g] (x) is not equal to [g o f] (x).

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- ShyLv 61 decade ago
Hi Z

1.

(x)=x+5

g(x)=x^2+6

(fog)x = (x^2 +6) +5 = x^2+11

(gof)x = (x+5)^2 + 6 = x^2 + 10x + 31

2.

f(x)=2x

g(x)=3x-4

(fog)x = 2(3x-4) = 6x-8

(gof)x = 3(2x) - 4 = 6x - 4

3.

f(x)=x-4

g(x)=3x^2

(fog)x = 3x^2 - 4

(gof)x = 3(x-4)^2 = 3x^2 - 24x + 48

Shy

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- Anonymous1 decade ago
1) fog(x)=f(g(x)),

let's say g(x)=x^2+6=y

so fog(x)=f((g(x))=f(y)=y+5

replace/substitute y=x^2+6

to get fog(x)= x^2+6+5=x^2+11

similarly gof(x)=g(f(x))

f(x)= x+5=y

resulting in gof(x)=g(y)=y^2+6 = (x+5)^2 +6

the others can be solved similarly

2)fog(x)=f(g(x))=f(3x-4)=2(3x-4)

gof(x)=g(f(x))=g(2x)=3(2x)-4

3)fog(x)=f(g(x))=f(3x^2)=3x^2-4

gof(x)=g(f(x))=g(x-4)=3(x-4)^2

Hope that helped

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- 1 decade ago
if u have (fog) put the g(x) equation in for the x in the f(x)

x^(2)+6+5=x^(2) + 11 that would be for (fog)

(x+5)^(2) +6= x^(2) +10x +31 (gof)

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- torneseLv 44 years ago
those definition of purposes inform you what comes out reckoning on what is going in. So utilising f(x)= 2x+5 Then f(sin(x)) = 2*sin(x) + 5 f( f(x) ) = f( 2x + 5) = 2*( 2x + 5) + 5 f (container) = 2*container + 5 and so on. Do you notice how this works?

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