Question

Are the functions f(x) = (x^2-1)/(x-1) and g(x)= x+1 equal for all x?

Answer 1

$\bf \textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\\\\ -------------------------------$

$\bf f(x)=\cfrac{x^2-1}{x-1}\implies f(x)=\cfrac{x^2-1^2}{x-1}\implies f(x)=\cfrac{(\underline{x-1})(x+1)}{\underline{x-1}} \\\\\\ f(x)=x+1\qquad \qquad \qquad \qquad \qquad g(x)=x+1\\\\ -------------------------------\\\\ \textit{they're, kinda, except that, when x = 1} \\\\\\ g(x)=(1)+1\implies g(x)=2 \\\\\\ f(x)=\cfrac{(1)^2-1}{(1)-1}\implies f(x)=\cfrac{0}{0}\impliedby und efined$