Answer
(a) $ 2x^2-x+1$
(b) $ \frac{2x^2-3x+2}{-2x+1}$
(c) $(-\infty, \frac{1}{2})\cup(\frac{1}{2},\infty)$
(d) $ 4x+2h-3$
(e) $ 0$
(f) $ -12$
(g) $ 1$
Work Step by Step
Given $f(x)=2x^2-3x+2$ and $g(x)=-2x+1$, we have:
(a) $(f-g)(x)=f(x)-g(x)=2x^2-3x+2+2x-1=2x^2-x+1$
(b) $(\frac{f}{g})(x)=\frac{f(x)}{g(x)}=\frac{2x^2-3x+2}{-2x+1}$
(c) The domain of $(\frac{f}{g})(x)$ is $x\ne\frac{1}{2}$ or $(-\infty, \frac{1}{2})\cup(\frac{1}{2},\infty)$
(d) $\frac{f(x+h)-f(x)}{h}=\frac{2(x+h)^2-3(x+h)+2-(2x^2-3x+2)}{h}=\frac{4xh+2h^2-3h}{h}=4x+2h-3$
(e) $(f+g)(1)=f(1)+g(1)=2(1)^2-3(1)+2-2(1)+1=0$
(f) $(fg)(2)=f(2)g(2)=(2(2)^2-3(2)+2)(-2(2)+1)=-12$
(g) $(f\circ g)(0)=(2(-2x+1)^2-3(-2x+1)+2)|_{x=0}=1$
