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)) ?

1 Answer
Aug 6, 2015

(i) True
(ii) False

Explanation:

f(x) is algebraic function sqrt(1–x) = (1–x)^(1/2) while g(x) 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\circg aka f(g(x)) to exist, the range of g, R_g (i.e. set of values of y when y = g(x)) 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(x)). That is, R_g \sube 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 = [1, \infty), 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 = RR_0^+ (that is, all non-negative real numbers) and R_g = RR (all real numbers).

In this case, R_f = RR_0^+ \sube (–\infty, 0) \cap (2, \infty) = D_g is not true, since RR_0^+ contains numbers that are not in (–\infty, 0) \cap (2, \infty) (e.g. [0, 2]), thus g\circf (i.e. g(f(x))) 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\circg aka f(g(x)) also does not exist.