Ok, I've literally just spent the last 2 hours just to figure out two compositions problems for homework, and I've about had it. Anyone here that can help?
Problem 1 $$ f(x) = 2x(2) - x -3 $$ $$ g(x) = x + 1 $$ Find: $$ f + g $$ $$ f - g $$ $$ f\cdot g $$ $$ \frac{f}{g} $$ $$ (f \circ g)(2) $$
Problem 2 $$ f(x) = 2x - 3 $$ $$ g(x) = \frac{x+3}{6} $$ Find: $$ (f\circ g)(x) $$ $$ (g\circ f)(x) $$ $$ (f\circ g)(2) $$ $$ (F\circ g)(6) $$
Some things to keep in mind:
$$\begin{align*} (f\pm g)(x)&=f(x)\pm g(x)\\\\ (f\cdot g)&=f(x)\cdot g(x)\\\\ \left(\frac{f}{g}\right)(x)&=\frac{f(x)}{g(x)}&(\text{provided that }g(x)\not=0)\\\\ (f\circ g)(x)&=f(g(x)) \end{align*}$$
You're given that $f(x)=2x^2-x-3$ and $g(x)=x+1$.
$$\begin{align*} (f+g)(x)&=(2x^2-x-3)+(x+1)\\\\ (f-g)(x)&=(2x^2-x-3)-(x+1)\\\\ (f\cdot g)&=(2x^2-x-3)(x+1)\\\\ \left(\frac{f}{g}\right)(x)&=\frac{2x^2-x-3}{x+1}\\\\ (f\circ g)(x)&=f(g(x))\\ &=2[g(x)]^2-g(x)-3\\ &=2(x+1)^2-(x+1)-3 \end{align*}$$