How to solve the function fg(2x)

I've become a bit confused on a question that is:

let $\displaystyle f(x)=x^2-1$ and $\displaystyle g(x)=2x+1$

Calculate $\displaystyle fg(2x)$

I'm not too sure. Is it:

$\displaystyle f(2(2x)+1)$ or $\displaystyle f(2(2x+1))$?

Thanks very much!

Re: How to solve the function fg(2x)

I think fg should be 2x^3 + x^2 - 2x - 1 and then you just plug in 2x in place of x to get your ans.

Re: How to solve the function fg(2x)

Quote:

Originally Posted by

**BobtheBob** I've become a bit confused on a question that is:

let $\displaystyle f(x)=x^2-1$ and $\displaystyle g(x)=2x+1$

Calculate $\displaystyle fg(2x)$

I'm not too sure. Is it:

$\displaystyle f(2(2x)+1)$ or $\displaystyle f(2(2x+1))$?

Thanks very much!

$\displaystyle (f \circ g)(x)=(2x+1)^2-1=4x^2+4x$

$\displaystyle (f \circ g) (2x)=4(2x)^2+4\cdot 2x=16x^2+8x$

Re: How to solve the function fg(2x)

Ok thanks very much for the help! Much appreciated.

Re: How to solve the function fg(2x)

Quote:

Originally Posted by

**princeps** $\displaystyle (f \circ g)(x)=(2x+1)^2-1=4x^2+4x$

$\displaystyle (f \circ g) (2x)=4(2x)^2+4\cdot 2x=16x^2+8x$

I think that is the wrong way of doing it. it's not fog

Re: How to solve the function fg(2x)

Oh right. I originaly got 16x^2 + 8x though although I wasn't sure of it...?