Question

The functions f and g are defined by f(x) =1/(1-3x) and g(x) =log_1/3(3x-2)-log₃(x) respectively

1. Write down the sets D_f (ehe domain of f) and D_g (the domain of g)

2. Solve the inequality f(x) > 2 for x$\isin$ D_f

3. Solve the inequality f(x) ≥ 2 for x$\isin$ D_g

Hint: Use the change of base formula

Answer 1

1.

$D_f:x\isin (-\infin,1/3)\cup (1/3,\infin)$

$g(x)=log_3(\frac{1}{x(3x-2)})$

$D_g:x\isin (-\infin,0)\cup (2/3,\infin)$

2.

$1/(1-3x)>2$ for $x\isin (-\infin,1/3)\cup (1/3,\infin)$

$0<1-3x<1/2$

$1/2<3x<1$

$1/6<x<1/3$

$x\isin (1/6,1/3)$

3.

$1/(1-3x)\ge2$ for $x\isin (-\infin,0)\cup (2/3,\infin)$

$0<1-3x\le1/2$

$1/2\le3x<1$

$1/6\le x<1/3$

$x\isin [1/6,1/3)$

no solutions because no intersection with D_g