Solving G(f(x)) - Help and Answers

[Deagonx]

Homework Statement

g(f(x))
g(x) = 3/x+1
f(x)= 3x+2

Homework Equations

?

The Attempt at a Solution

Ive had none I don't even know how to attempt this problem.

 

[dynamicsolo]

When you see a composite function, say, g( f(x) ) , that is saying that we will substitute the function f(x) everywhere that 'x' appears in the expression for g(x) . In your problem, this will mean that if g(x) = 3/x+1 ,

by the way, is this [tex]\frac{3}{x+1} or \frac{3}{x} + 1 [/tex] (I will assume the first -- if you don't use TeX, be sure to use parentheses)

you would start with

g( f(x) ) = 3 / [ f(x) + 1 ] = 3 / [ ( 3x + 2 ) + 1 ] ,

and make any necessary algebraic simplifications from there.

 

[GreenPrint]

every were you see x in g(x) plug in f(x)

 

[I like Serena]

Welcome to PF, Deagonx!

Try it like this:
g(f(x))
g(u) = 3/u+1
u = f(x) = 3x+2

 

[Ray Vickson]

Homework Statement

g(f(x))
g(x) = 3/x+1
f(x)= 3x+2

Homework Equations

?

The Attempt at a Solution

Ive had none I don't even know how to attempt this problem.

Maybe you know whether g(x) = 3/(x+1) or g(x) = (3/x) + 1, but nobody else does. Which one do you mean? (If I read it according to *standard rules*, it means the second form.)

RGV

 

[Deagonx]

They way that I did it was as followed:

If f(x) = 3x + 2. and the equation is g(f(x)) then isn't it really just g(3x+2)

In which case, I came to 3/(3x+2) + 1 which I then simplified to x + 2.5 Which I think is wrong.

 

[dynamicsolo]

They way that I did it was as followed:

If f(x) = 3x + 2. and the equation is g(f(x)) then isn't it really just g(3x+2)

This part is correct.

In which case, I came to 3/(3x+2) + 1 which I then simplified to x + 2.5 Which I think is wrong.

So [tex] f( g(x) ) = \frac{3}{(3x+2) + 1 } = \frac{3}{3x + 3} .[/tex] What would that simplify to?

 

[Deagonx]

This part is correct.



So [tex] f( g(x) ) = \frac{3}{(3x+2) + 1 } = \frac{3}{3x + 3} .[/tex] What would that simplify to?

x + 1?

 

[dynamicsolo]

This is a ratio: what can be done in the numerator and denominator? (Remember, the 3x + 3 is in the denominator.)

 

[Mentallic]

And remember, if the original expression is equal to your new expression, whatever value of x you choose (granted you don't divide by zero), you should be able to plug it into both expressions and the same answer will come out!

 

[uart]

They way that I did it was as followed:

If f(x) = 3x + 2. and the equation is g(f(x)) then isn't it really just g(3x+2)

In which case, I came to 3/(3x+2) + 1 which I then simplified to x + 2.5 Which I think is wrong.

The first part was correct, but the simplification was wrong.

To simplify that expression first put everything on a common denominator.

[tex]\frac{3}{3x+2} + 1 = \frac{3}{3x+2} + \frac{3x+2}{3x+2} [/tex]

Then simplify it from there.

 

[dynamicsolo]

Evidently, everyone still hasn't settled on what g(x) is. So at least we'll have both possible versions...

By the way, in this version,
[tex] f( g(x) ) = \frac{3}{(3x+2) + 1 } = \frac{3}{3x + 3} ,[/tex]

you are dividing 3 by ( 3x + 3 ) , not ( 3x + 3 ) by 3 : that's why this version isn't x + 1 .

 

[uart]

Evidently, everyone still hasn't settled on what g(x) is ...

Yes it's difficult when the OP won't clarify that, even after being prompted a few times.

But without any other clarification I think we really have to go with what the OP is typing, whether or not that is really what he/she is actually trying to ask.