Functions and Rearranging Help

[liamporter1702]

Two functions are defined as:
f(x) = 1/(-10x+14)+17
g(x) = 1/(-4x-3)

Find fg(x) and gf(x) and simplify answer into form (ax+b)/(cx+d) where a,b,c and d are numbers to be found.

I know it is common practice to put in some of my own attempts onto here, but I am not even getting remotely close to an answer. :s

Any help or guidance would be greatly appreciated!

 

[GFauxPas]

fg(x) is defined to be equal to f(x)g(x) for all x in the intersection of the domains of f and g, so you just multiply the equations together.

But it sounds like you are expecting fg and gf to be two different functions, which suggests there's a typo in your question.

edit: there has to be a typo in your question, because fg and gf wouldn't be of the form (ax+b)/(cx+d)

 

[Mark44]

Two functions are defined as:
f(x) = 1/(-10x+14)+17
g(x) = 1/(-4x-3)

For the first function, what you wrote is
$$ f(x) = \frac{1}{-10x + 14} + 17$$
Is that what you intended to write?

Find fg(x) and gf(x) and simplify answer into form (ax+b)/(cx+d) where a,b,c and d are numbers to be found.

I know it is common practice to put in some of my own attempts onto here, but I am not even getting remotely close to an answer. :s

Any help or guidance would be greatly appreciated!

 

[Dick]

fg(x) is defined to be equal to f(x)g(x) for all x in the intersection of the domains of f and g, so you just multiply the equations together.

But it sounds like you are expecting fg and gf to be two different functions, which suggests there's a typo in your question.

edit: there has to be a typo in your question, because fg and gf wouldn't be of the form (ax+b)/(cx+d)

I think the question is to evaluate the compositions, not the product. I.e. f(g(x)) and g(f(x)). Those are different and are of that form.

 

[Mark44]

I think the question is to evaluate the compositions, not the product. I.e. f(g(x)) and g(f(x)). Those are different and are of that form.

That's my take as well, Dick.

 

[jtbell]

I know it is common practice to put in some of my own attempts onto here, but I am not even getting remotely close to an answer.

At least show us how you started. Then we can tell you if you're starting off OK, and if not, give you hints on which way you should be going.

 

[Mark44]

I know it is common practice to put in some of my own attempts onto here, but I am not even getting remotely close to an answer. :s

It's not just common practice - it's a requirement of this board.

 

[liamporter1702]

From what I understood of the question and from the previous questions is to substitute g(x) into f(x), which I think you're right, it would be written like f(g(x)) and vice versa for g(f(x)). Sorry about not making this clearer, this is my first time learning about functions.

I attempted to substitute 1/(-4x-3) into 1/(-10x+14)+17 to give 1/(-10(1/(-4x-3))+14)+17 (sorry I can't make these equations easier to read, I'm not sure how to).

From there on I got a bit lost with the simplifying.

 

[liamporter1702]

Am I on the right track by substituting one equation into the other and and I just need to look back at my simplifying or am I going about this completely wrong?

 

[Mark44]

Am I on the right track by substituting one equation into the other and and I just need to look back at my simplifying or am I going about this completely wrong?

What you did in your previous post looks like you're on the right track.

 

[RUber]

You have 1/(-10(1/(-4x-3))+14)+17 = f(g(x)).
To simplify, $\frac{1}{-10 \left(\frac{1}{-4x-3} \right)+14}+17$, first, combine terms on the bottom of the fraction to get something of the form $\frac{1}{\frac{ax+b}{cx+d}}+17$
Next, note that $\frac{1}{\frac{ax+b}{cx+d}}=\frac{cx+d}{ax+b}$

 

[GFauxPas]

From what I understood of the question and from the previous questions is to substitute g(x) into f(x), which I think you're right, it would be written like f(g(x)) and vice versa for g(f(x)). Sorry about not making this clearer, this is my first time learning about functions

You can also write $f \circ g (x) = f(g(x)), g\circ f (x) = g(f(x))$ if you're looking for notation.