1. ## Help woth functions

Hello!

I need help to solve this:

If F( ( x + 3 ) / ( x + 1 ) ) = 3x + 2 x belongs to R \ {!-1}
then F( 5 ) = ?

It is a first time ever for me to have met this kind of problems and functions in general (I'm trying to teach myself math to get on university), I have very weak background in math so you can imagine my struggle.

I tried to do something like this:

x = 1
F( ( 1 + 3 ) / ( 1 + 1 ) ) = 3 * 1 + 2
F( 4 / 2 ) = 5

so, I have now 4/2 and 5 and I say F(4/2) = 5
and to find F(5) I did this: first I found by how much I need to multiply 4/2 to get 5:
4/2 * y = 5
y = 5 * (2/4)
y = 10/4
y = 5/2

now, I know that in order to get the value of F(5) I need to divide y with f(4/2):
f(4/2) = 5
so, y / 5 = 5/2 * 1/5 = 1/2 --- and the book states that is the right answer, BUT when I used for x some other number, like 3, I didn't get that answer. Please help.

2. Solve $\displaystyle \frac{x+3}{x+1} = 5$

3. They try to trick you with f(x)= bla bla.
In general, it's helpful to read: f(x) as "f function of x".
Hence, y(x) as "y function of x".

So, in your case drop that F thing completely.
(x+3)/(x+1) = some constant

4. Originally Posted by pickslides
Solve $\displaystyle \frac{x+3}{x+1} = 5$
Ugh... I don't get it...

I solved and got 1/2, and thats the right answer, but...

I mean, what's the purpose of the right side of equation then (3x + 2)?

5. Originally Posted by ikislav
Ugh... I don't get it...

I solved and got 1/2, and thats the right answer, but...

I mean, what's the purpose of the right side of equation then (3x + 2)?
You are looking for F(5) . The number 5 tells you what x is, then the 3x + 2 gives you the value for F(5).

-Dan

If this is really giving you problems, there is a way to put F into a more standard form.

Let $\displaystyle y = \frac{x + 3}{x + 1}$

Solve for x:
$\displaystyle x = - \frac{y - 3}{y - 1}$

and put this in for F():
$\displaystyle F(y) = -3 \left ( \frac{y - 3}{y - 1} \right ) + 2$

and solve your problem by putting y = 5.