Question

Someone help me with the 2 problems I posted yesterday and didn't get answers to... (they're on my profile)

Answer 1

Ykz can you help or no?

Answer 2

I get the idea of plugging in one for the other but the problems look weird because they are fractional.

Answer 3

phi can u help?

Answer 4

\[f \circ g= f(g(x))\]\[g(x)= \frac{-1-5x}{5x}\]\[f(x)=\frac{-1}{5x+5}\]

Answer 5

so far so good

Answer 6

get on the chat thing Foster

Answer 7

The easiest way to do this is concentrate on just the denominator 5x+5 in f(x)
replace the x with g(x):
\[5\cdot \frac{-1-5x}{5x}+5 \]
simplify by canceling the 5's. Also, the make a common denominator for the second 5:
\[\frac{-1-5x}{x}+\frac{5}{1}\cdot \frac{x}{x}= \frac{-1-5x+5x}{x}= \frac{-1}{x}\]

Answer 8

Now we have the denominator of f(g(x)) simplified to -1/x. So the total expression is
\[\frac{-1}{\frac{-1}{x}}= -1\cdot \frac{x}{-1}=x\]

Answer 9

im confused

Answer 10

shouldnt f(g(x)) = f(-1 / 5(-1-5x/5x)+5)

Answer 11

when u plug in one for the other

Answer 12

what f(g(x)) means is look at the f function, and everywhere you see x, replace it with g(x)
so, without expanding it, we would get:
\[f(g(x))= \frac{-1}{5g(x)+5}\]
of course g(x) is a messy fraction, but you get the idea?

Answer 13

yeah i know that i have to plug in the fraction for g(x)... just need help simplifying

Answer 14

so that i get x from there

Answer 15

OK, first simplify 5g(x)

Answer 16

do i multiply 5 on the bottom and top of the fraction?

Answer 17

so i would get -5 -25x / 25x

Answer 18

\[5g(x)= 5\cdot \frac{-1-5x}{5x}\]

Answer 19

So the 5 in top and the bottom cancel each other out (5/5 = 1)

Answer 20

so i dont even have to multiply cause they cancel out?

Answer 21

If you are following, then you get
\[\frac{-1-5x}{x} \]

Answer 22

Now add 5.

Answer 23

foster: delete ur comments and it wont show up

Answer 24

It is always good to cancel things. You could multiply things out but it's messier

Answer 25

yeah i agree

Answer 26

let me do it on paper and ill get back to you in a sec or 2

Answer 27

ok and then?

Answer 28

You asked above: so i would get -5 -25x / 25x

no. when you multiply fractions, you multiply top times top, bottom times bottom
In the case of a whole number like 5, you assume the bottom is 1 (5/1 = 5).
So, if you did multiply things out it would be
\[\frac{5}{1}\cdot\frac{-1-5x}{5x}= \frac{-5-25x}{5x}\]

Answer 29

yeah i get that

Answer 30

Did you add 5 to
\[\frac{-1-5x}{x} \]
?

Answer 31

too many rules to remember sometimes

Answer 32

I agree about the rules. lots of them to remember.

Answer 33

yes and got -1 / 4-5x/x

Answer 34

You got
\[ \frac{-1}{4} - \frac{5x}{x} \]

Answer 35

How?!

Answer 36

i added 5 to -1 so i got 4

Answer 37

OK, some more rules.
http://www.khanacademy.org/video/adding-fractions-with-like-denominators?playlist=Developmental+Math
http://www.khanacademy.org/video/adding-fractions-with-unlike-denominators?playlist=Developmental+Math
when you have time. They are very short, but helpful

Answer 38

why? what am i doing wrong?

Answer 39

Meanwhile, to add fractions, you make sure both fractions have the same bottom, and then you can add the tops. 5 is a whole number, so its bottom is understood to be 1

Answer 40

The first fraction has an x in the bottom. So you have to change the 5 to have an x in its bottom.

Answer 41

o

Answer 42

so the same fractional rules apply with variables as well

Answer 43

The upside to the rules, no exceptions.

Answer 44

i see

Answer 45

i got x :D

Answer 46

when you add 5, you change it to 5x/x and now you add the tops
(-1-5x+5x)/x = -1/x
so you should get -1/x