Question

Let $f:R\to R:f\left(x\right)=x+2$ and $g:R-\left\{2\right\}\to R:g\left(x\right)=\frac{x^2-4}{x-2}$
Show that $f\ne g$. Re-define f and g such that f = g

Answer 1

We have, $\text{dom (f)}=R$ and $\text{dom (g)}=R-\left\{2\right\}$

Since $\text{dom (f)}\ne \text{dom (g)}$, so we have $f\ne g$.

For every real number $x\ne 2$, we have

$g\left(x\right)=\frac{x^2-4}{x-2}=\frac{(x-2)(x+2)}{(x-2)}=(x+2)$ $[\because (x-2)\ne 0]$.

Thus, $f\left(x\right)=g\left(x\right)$ for all $xϵR-\left\{2\right\}$

$\therefore$ f = g only when, they are re-defined as under.

$f:R-\left\{2\right\}\to R:f\left(x\right)=x+2$ and $g:R-\left\{2\right\}\to R:g\left(x\right)=\frac{x^2-4}{x-2}$