Question

for f(x)=1/x-5 and g(x)=x^2+2, find:
(fog)(x) and (gof)(6)

Answer 1

\[(f\circ g)(x)=f(g(x))=\frac{1}{g(x)-5}=\frac{1}{(x^2+2)-5}=\frac{1}{x^2-3}\]
Do something similar for \((g\circ f(x)\)).

Answer 2

how did you do it?

Answer 3

(f o g)(x) =
you plug in what g(x) is = to into the equation for f(x)

Answer 4

if f(x) = 2x
and g(x) = -2
(f o g)(x) = 2(-2)
understand?..

Answer 5

(g o f)(6)
simply means you evaluate f(6)
and then put that result into the equation for g(x).

Answer 6

Okay i'm getting it. It would be 1/x^2-3, but what about the second one i'm even more confused.

Answer 7

f(x)=1/x-5
First, find f(6)

Answer 8

To start off the second problem, can you find f(6)?

Answer 9

I have no idea :/

Answer 10

f(6) means you substitute for all values of x.
f(x) = 1/x-5
f(6) = 1/6-5 = 1/1 = 1
Understand?

Answer 11

Okay

Answer 12

From there, you plug in f(6)
into g(x)
g(x)=x^2+2
(g o f)(6) = 1^2 + 2
= 3

Answer 13

Ok thank you

Answer 14

so 2 is (g o f)(6) = 1^2 + 2
and 1 is f(x)=1/x-5

Answer 15

number 1 is
(g o f) as the zzrockr said in his explanation it is 1/x^2-3

Answer 16

(g o f)(6) = g(f(6))

Find f(6) first.
f(x) = 1/(x - 5)
f(6) = 1/(6 - 5) = 1/1 = 1

g(f(6)) = g(1)
g(x) = x^2 + 2
g(1) = 1^2 + 2 = 1 + 2 = 3

Answer 17

For the first part:
If (f o g)(6) = f(g(6))

Find g(6) first:
g(x) = x^2 + 2
g(6) = 6^2 + 2 = 36 + 2 = 38

f(g(6)) = f(38)
f(x) = 1/(x - 5)
f(38) = 1/ (38 - 5) = 1/33