Given $$f (x) = x - 1$$ and $$g (x) =\dfrac{x^2}{1-x}$$, find $$(f - g) (x), (f \cdot g) (x)$$.
See development and solution
Development:
$$(f-g) (x) = f (x) - g (x) = x-1-\dfrac{x^2}{1-x}=\dfrac{-x^2+2x-1-x^2}{1-x}=$$
$$=\dfrac{2x^2-2x+1}{x-1}$$
$$(f\cdot g)(x)=f(x)\cdot g(x)=(x-1)\cdot \dfrac{x^2}{1-x}=(-1)(1-x)\dfrac{x^2}{1-x}=-x^2$$
Solution:
$$(f-g) (x) = \dfrac{2x^2-2x+1}{x-1}$$
$$(f\cdot g)(x)=-x^2$$