Rational Derivative

Let $g(x)=x^2+x+2$ . We have $f(x)=g(x)$ for any $x\neq 0$ ; specifically at and around $x=1$ , meaning that $f'(1)-3=g'(1)-3=2\cdot 1+1-3=\boxed{0}$ .

Let g(x) = x^{3} + x^{2} + 2x and h(x) = x f'(x) = $\frac{g'(x)h(x) - h'(x)g(x)}{((h(x))^{2}}$ g'(x) = 3x^{2} + 2x +2 h'(x) = 1 f'(x) = $\frac{(3x^{2}+2x+2)(x) - (1)(x^{3}+x^{2}+2x)}{x^{2}}$ f'(x) = $\frac{(3x^{2}+2x+2)(x)-(x^{3}+x^{2}+2x}{x^{2}}$ f'(1) = $\frac{(3(1)^{2}+2(1)+2)(1)-(1^{3}+1^{2}+2}{1^{2}}$ = $\frac{(3+2+2)(1)-(1+1+2)}{1}$ = $\frac{7-4}{1}$ = 3 f'(1) - 3 = 3 -3 = 0

$f(x) = \dfrac{x^3 + 2x + x^2}{x} = x^2 + 2 + x \\ f'(x) = 2x + 1 \\ f'(1) = 3 \\ \therefore f'(1) - 3 = \boxed0$