# F(x)=-√(1-x) and g(x)=ln x (x-2). TRUE OR FALSE.explain. (i)none algebraic operation between f and g. (ii)g(f(x)) exists but not f(g(x)) ?

Aug 6, 2015

(i) True
(ii) False

#### Explanation:

$f \left(x\right)$ is algebraic function sqrt(1–x) = (1–x)^(1/2) while $g \left(x\right)$ is transcendental function (logarithmic), since algebraic operations transform algebraic expressions into algebraic expressions and transcendental expressions into transcendental expressions but not algebraic expressions into transcendental expressions.

For a composite function $f \setminus \circ g$ aka $f \left(g \left(x\right)\right)$ to exist, the range of $g$, ${R}_{g}$ (i.e. set of values of $y$ when $y = g \left(x\right)$) must be smaller than or equal to (i.e. a subset) the domain of $f$, ${D}_{f}$ (i.e. the $x$ values that are 'plugged' into the equation $y = f \left(x\right)$). That is, ${R}_{g} \setminus \subseteq {D}_{f}$.

Since the question does not state the domain or range of $f$ (i.e. ${D}_{f}$) or $g$ (i.e. ${D}_{g}$), I will assume that ${D}_{f} = \left[1 , \setminus \infty\right)$, that is, all real numbers greater than or equal to 1 and D_g = (–\infty, 0) \cap (2, \infty), that is, all real numbers small than 0 or greater than 2 (to allow whatever is in the $\ln$ function to be positive). Thus, plugging the sets of values of $x$ into $f$ and $g$ produces ${R}_{f} = {\mathbb{R}}_{0}^{+}$ (that is, all non-negative real numbers) and ${R}_{g} = \mathbb{R}$ (all real numbers).

In this case, R_f = RR_0^+ \sube (–\infty, 0) \cap (2, \infty) = D_g is not true, since ${\mathbb{R}}_{0}^{+}$ contains numbers that are not in (–\infty, 0) \cap (2, \infty) (e.g. $\left[0 , 2\right]$), thus $g \setminus \circ f$ (i.e. $g \left(f \left(x\right)\right)$) does not exist. R_g = RR \sube [1,∞) = D_f also doesn't hold as it includes elements absent in ${D}_{f}$ (i.e. (–\infty, 1)), thus $f \setminus \circ g$ aka $f \left(g \left(x\right)\right)$ also does not exist.